Trigonometric equations

Answer: $x = \frac{3\pi}{4} \text{ or } x = \frac{5\pi}{4}$

Divide by $2\mathclose{:}$ $\cos(x) = - \frac{\sqrt{2}}{2}\mathclose{.}$ One angle is $\alpha = \arccos\left(- \frac{\sqrt{2}}{2}\right) = \frac{3\pi}{4}\mathclose{;}$ cosine is also equal at $-\alpha\mathclose{,}$ which is $2\pi - \alpha$ in one turn. So $x = \frac{3\pi}{4} \text{ or } x = \frac{5\pi}{4}\mathclose{.}$

Common mistakes, and the answer each one gives:

• Took the second solution as $\pi - \alpha\mathclose{,}$ the rule for sine → $x = \frac{\pi}{4} \text{ or } x = \frac{3\pi}{4}$
• Gave only the first solution → $x = \frac{3\pi}{4}$

Answer: $x = \frac{\pi}{6} \text{ or } x = \frac{\pi}{3}$

Divide by $5\mathclose{:}$ $\sin(2x) = \frac{\sqrt{3}}{2}\mathclose{.}$ Put $u = 2x\mathclose{,}$ with $0 \le u < 2\pi\mathclose{:}$ $\sin u = \frac{\sqrt{3}}{2}$ gives $u = \alpha$ or $u = \pi - \alpha$ with $\alpha = \frac{\pi}{3}\mathclose{.}$ Halve each to get $x\mathclose{.}$ So $x = \frac{\pi}{6} \text{ or } x = \frac{\pi}{3}\mathclose{.}$

Common mistakes, and the answer each one gives:

• Solved for $2x$ and forgot to halve → $x = \frac{\pi}{3} \text{ or } x = \frac{2\pi}{3}$
• Gave only the first solution → $x = \frac{\pi}{6}$

Answer: $x = \frac{2\pi}{3} \text{ or } x = \frac{4\pi}{3}$

Put $u = \cos(x)\mathclose{:}$ $2 u^{2} + 5 u + 2 = 0\mathclose{,}$ so $u = -2$ or $u = - \frac{1}{2}\mathclose{.}$ A cosine lies between $-1$ and $1\mathclose{,}$ so $\cos(x) = -2$ has no solution. Every angle in one turn with $\cos(x)$ equal to $- \frac{1}{2}\mathclose{:}$ $x = \frac{2\pi}{3} \text{ or } x = \frac{4\pi}{3}\mathclose{.}$

Common mistakes, and the answer each one gives:

• Took only the first angle for each value → $x = \frac{2\pi}{3}$

Answer: $x = \frac{5\pi}{4} \text{ or } x = \frac{7\pi}{4}$

Divide by $5\mathclose{:}$ $\sin(x) = - \frac{\sqrt{2}}{2}\mathclose{.}$ One angle is $\alpha = \arcsin\left(- \frac{\sqrt{2}}{2}\right) = -\frac{\pi}{4}\mathclose{;}$ sine is also equal at $\pi - \alpha\mathclose{,}$ and both are brought into $[0, 2\pi)\mathclose{.}$ So $x = \frac{5\pi}{4} \text{ or } x = \frac{7\pi}{4}\mathclose{.}$

Common mistakes, and the answer each one gives:

• Took the second solution as $\pi + \alpha$ instead of $\pi - \alpha$ → $x = \frac{3\pi}{4} \text{ or } x = \frac{7\pi}{4}$
• Gave only the first solution → $x = \frac{5\pi}{4}$

Answer: $x = \frac{3\pi}{4} \text{ or } x = \frac{7\pi}{4}$

Divide by $5\mathclose{:}$ $\tan(x) = -1\mathclose{.}$ One angle is $\arctan\left(-1\right) = -\frac{\pi}{4}\mathclose{,}$ and tangent repeats every $\pi\mathclose{.}$ So $x = \frac{3\pi}{4} \text{ or } x = \frac{7\pi}{4}\mathclose{.}$

Common mistakes, and the answer each one gives:

• Gave one solution; tangent repeats every $\pi\mathclose{,}$ so there are two in one turn → $x = \frac{3\pi}{4}$

Answer: $x = \frac{5\pi}{6} \text{ or } x = \frac{7\pi}{6}$

Divide by $6\mathclose{:}$ $\cos(x) = - \frac{\sqrt{3}}{2}\mathclose{.}$ One angle is $\alpha = \arccos\left(- \frac{\sqrt{3}}{2}\right) = \frac{5\pi}{6}\mathclose{;}$ cosine is also equal at $-\alpha\mathclose{,}$ which is $2\pi - \alpha$ in one turn. So $x = \frac{5\pi}{6} \text{ or } x = \frac{7\pi}{6}\mathclose{.}$

Common mistakes, and the answer each one gives:

• Took the second solution as $\pi - \alpha\mathclose{,}$ the rule for sine → $x = \frac{\pi}{6} \text{ or } x = \frac{5\pi}{6}$
• Gave only the first solution → $x = \frac{5\pi}{6}$

Answer: $x = \frac{\pi}{12} \text{ or } x = \frac{5\pi}{12}$

Divide by $5\mathclose{:}$ $\sin(2x) = \frac{1}{2}\mathclose{.}$ Put $u = 2x\mathclose{,}$ with $0 \le u < 2\pi\mathclose{:}$ $\sin u = \frac{1}{2}$ gives $u = \alpha$ or $u = \pi - \alpha$ with $\alpha = \frac{\pi}{6}\mathclose{.}$ Halve each to get $x\mathclose{.}$ So $x = \frac{\pi}{12} \text{ or } x = \frac{5\pi}{12}\mathclose{.}$

Common mistakes, and the answer each one gives:

• Solved for $2x$ and forgot to halve → $x = \frac{\pi}{6} \text{ or } x = \frac{5\pi}{6}$
• Gave only the first solution → $x = \frac{\pi}{12}$

Answer: $x = 0,\ x = \pi,\ x = \frac{7\pi}{6} \text{ or } x = \frac{11\pi}{6}$

Put $u = \sin(x)\mathclose{:}$ $2 u^{2} + u = 0\mathclose{,}$ so $u = - \frac{1}{2}$ or $u = 0\mathclose{.}$ Every angle in one turn with $\sin(x)$ equal to $- \frac{1}{2}$ or $0\mathclose{:}$ $x = 0,\ x = \pi,\ x = \frac{7\pi}{6} \text{ or } x = \frac{11\pi}{6}\mathclose{.}$

Common mistakes, and the answer each one gives:

• Took only the first angle for each value → $x = 0 \text{ or } x = \frac{11\pi}{6}$
• Solved only the factor that gives $\sin(x) = - \frac{1}{2}$ → $x = \frac{7\pi}{6} \text{ or } x = \frac{11\pi}{6}$

Answer: $x = \frac{7\pi}{6} \text{ or } x = \frac{11\pi}{6}$

Divide by $4\mathclose{:}$ $\sin(x) = - \frac{1}{2}\mathclose{.}$ One angle is $\alpha = \arcsin\left(- \frac{1}{2}\right) = -\frac{\pi}{6}\mathclose{;}$ sine is also equal at $\pi - \alpha\mathclose{,}$ and both are brought into $[0, 2\pi)\mathclose{.}$ So $x = \frac{7\pi}{6} \text{ or } x = \frac{11\pi}{6}\mathclose{.}$

Common mistakes, and the answer each one gives:

• Took the second solution as $\pi + \alpha$ instead of $\pi - \alpha$ → $x = \frac{5\pi}{6} \text{ or } x = \frac{11\pi}{6}$
• Gave only the first solution → $x = \frac{7\pi}{6}$

Answer: $x = \frac{3\pi}{4} \text{ or } x = \frac{7\pi}{4}$

Divide by $3\mathclose{:}$ $\tan(x) = -1\mathclose{.}$ One angle is $\arctan\left(-1\right) = -\frac{\pi}{4}\mathclose{,}$ and tangent repeats every $\pi\mathclose{.}$ So $x = \frac{3\pi}{4} \text{ or } x = \frac{7\pi}{4}\mathclose{.}$

Common mistakes, and the answer each one gives:

• Gave one solution; tangent repeats every $\pi\mathclose{,}$ so there are two in one turn → $x = \frac{3\pi}{4}$

Answer: $x = \frac{\pi}{4} \text{ or } x = \frac{7\pi}{4}$

Divide by $2\mathclose{:}$ $\cos(x) = \frac{\sqrt{2}}{2}\mathclose{.}$ One angle is $\alpha = \arccos\left(\frac{\sqrt{2}}{2}\right) = \frac{\pi}{4}\mathclose{;}$ cosine is also equal at $-\alpha\mathclose{,}$ which is $2\pi - \alpha$ in one turn. So $x = \frac{\pi}{4} \text{ or } x = \frac{7\pi}{4}\mathclose{.}$

Common mistakes, and the answer each one gives:

• Took the second solution as $\pi - \alpha\mathclose{,}$ the rule for sine → $x = \frac{\pi}{4} \text{ or } x = \frac{3\pi}{4}$
• Gave only the first solution → $x = \frac{\pi}{4}$

Answer: $x = \frac{7\pi}{12} \text{ or } x = \frac{11\pi}{12}$

Divide by $3\mathclose{:}$ $\sin(2x) = - \frac{1}{2}\mathclose{.}$ Put $u = 2x\mathclose{,}$ with $0 \le u < 2\pi\mathclose{:}$ $\sin u = - \frac{1}{2}$ gives $u = \alpha$ or $u = \pi - \alpha$ with $\alpha = -\frac{\pi}{6}\mathclose{,}$ which is $\frac{11\pi}{6}$ in $[0, 2\pi)\mathclose{.}$ Halve each to get $x\mathclose{.}$ So $x = \frac{7\pi}{12} \text{ or } x = \frac{11\pi}{12}\mathclose{.}$

Common mistakes, and the answer each one gives:

• Solved for $2x$ and forgot to halve → $x = \frac{7\pi}{6} \text{ or } x = \frac{11\pi}{6}$
• Gave only the first solution → $x = \frac{7\pi}{12}$

Answer: $x = \frac{\pi}{2} \text{ or } x = \frac{3\pi}{2}$

Put $u = \cos(x)\mathclose{:}$ $u^{2} - 2 u = 0\mathclose{,}$ so $u = 0$ or $u = 2\mathclose{.}$ A cosine lies between $-1$ and $1\mathclose{,}$ so $\cos(x) = 2$ has no solution. Every angle in one turn with $\cos(x)$ equal to $0\mathclose{:}$ $x = \frac{\pi}{2} \text{ or } x = \frac{3\pi}{2}\mathclose{.}$

Common mistakes, and the answer each one gives:

• Took only the first angle for each value → $x = \frac{\pi}{2}$

Answer: $x = \frac{\pi}{6} \text{ or } x = \frac{5\pi}{6}$

Divide by $4\mathclose{:}$ $\sin(x) = \frac{1}{2}\mathclose{.}$ One angle is $\alpha = \arcsin\left(\frac{1}{2}\right) = \frac{\pi}{6}\mathclose{;}$ sine is also equal at $\pi - \alpha\mathclose{,}$ and both are brought into $[0, 2\pi)\mathclose{.}$ So $x = \frac{\pi}{6} \text{ or } x = \frac{5\pi}{6}\mathclose{.}$

Common mistakes, and the answer each one gives:

• Took the second solution as $\pi + \alpha$ instead of $\pi - \alpha$ → $x = \frac{\pi}{6} \text{ or } x = \frac{7\pi}{6}$
• Gave only the first solution → $x = \frac{\pi}{6}$

Answer: $x = \frac{2\pi}{3} \text{ or } x = \frac{5\pi}{3}$

Divide by $4\mathclose{:}$ $\tan(x) = - \sqrt{3}\mathclose{.}$ One angle is $\arctan\left(- \sqrt{3}\right) = -\frac{\pi}{3}\mathclose{,}$ and tangent repeats every $\pi\mathclose{.}$ So $x = \frac{2\pi}{3} \text{ or } x = \frac{5\pi}{3}\mathclose{.}$

Common mistakes, and the answer each one gives:

• Gave one solution; tangent repeats every $\pi\mathclose{,}$ so there are two in one turn → $x = \frac{2\pi}{3}$

Answer: $x = \frac{\pi}{4} \text{ or } x = \frac{7\pi}{4}$

Divide by $3\mathclose{:}$ $\cos(x) = \frac{\sqrt{2}}{2}\mathclose{.}$ One angle is $\alpha = \arccos\left(\frac{\sqrt{2}}{2}\right) = \frac{\pi}{4}\mathclose{;}$ cosine is also equal at $-\alpha\mathclose{,}$ which is $2\pi - \alpha$ in one turn. So $x = \frac{\pi}{4} \text{ or } x = \frac{7\pi}{4}\mathclose{.}$

Common mistakes, and the answer each one gives:

• Took the second solution as $\pi - \alpha\mathclose{,}$ the rule for sine → $x = \frac{\pi}{4} \text{ or } x = \frac{3\pi}{4}$
• Gave only the first solution → $x = \frac{\pi}{4}$

Answer: $x = \frac{2\pi}{3} \text{ or } x = \frac{5\pi}{6}$

Divide by $4\mathclose{:}$ $\sin(2x) = - \frac{\sqrt{3}}{2}\mathclose{.}$ Put $u = 2x\mathclose{,}$ with $0 \le u < 2\pi\mathclose{:}$ $\sin u = - \frac{\sqrt{3}}{2}$ gives $u = \alpha$ or $u = \pi - \alpha$ with $\alpha = -\frac{\pi}{3}\mathclose{,}$ which is $\frac{5\pi}{3}$ in $[0, 2\pi)\mathclose{.}$ Halve each to get $x\mathclose{.}$ So $x = \frac{2\pi}{3} \text{ or } x = \frac{5\pi}{6}\mathclose{.}$

Common mistakes, and the answer each one gives:

• Solved for $2x$ and forgot to halve → $x = \frac{4\pi}{3} \text{ or } x = \frac{5\pi}{3}$
• Gave only the first solution → $x = \frac{2\pi}{3}$

Answer: $x = \frac{\pi}{6} \text{ or } x = \frac{5\pi}{6}$

Put $u = \sin(x)\mathclose{:}$ $2 u^{2} + 3 u - 2 = 0\mathclose{,}$ so $u = \frac{1}{2}$ or $u = -2\mathclose{.}$ A sine lies between $-1$ and $1\mathclose{,}$ so $\sin(x) = -2$ has no solution. Every angle in one turn with $\sin(x)$ equal to $\frac{1}{2}\mathclose{:}$ $x = \frac{\pi}{6} \text{ or } x = \frac{5\pi}{6}\mathclose{.}$

Common mistakes, and the answer each one gives:

• Took only the first angle for each value → $x = \frac{\pi}{6}$

Answer: $x = \frac{4\pi}{3} \text{ or } x = \frac{5\pi}{3}$

Divide by $3\mathclose{:}$ $\sin(x) = - \frac{\sqrt{3}}{2}\mathclose{.}$ One angle is $\alpha = \arcsin\left(- \frac{\sqrt{3}}{2}\right) = -\frac{\pi}{3}\mathclose{;}$ sine is also equal at $\pi - \alpha\mathclose{,}$ and both are brought into $[0, 2\pi)\mathclose{.}$ So $x = \frac{4\pi}{3} \text{ or } x = \frac{5\pi}{3}\mathclose{.}$

Common mistakes, and the answer each one gives:

• Took the second solution as $\pi + \alpha$ instead of $\pi - \alpha$ → $x = \frac{2\pi}{3} \text{ or } x = \frac{5\pi}{3}$
• Gave only the first solution → $x = \frac{4\pi}{3}$

Answer: $x = \frac{\pi}{6} \text{ or } x = \frac{7\pi}{6}$

Divide by $3\mathclose{:}$ $\tan(x) = \frac{\sqrt{3}}{3}\mathclose{.}$ One angle is $\arctan\left(\frac{\sqrt{3}}{3}\right) = \frac{\pi}{6}\mathclose{,}$ and tangent repeats every $\pi\mathclose{.}$ So $x = \frac{\pi}{6} \text{ or } x = \frac{7\pi}{6}\mathclose{.}$

Common mistakes, and the answer each one gives:

• Gave one solution; tangent repeats every $\pi\mathclose{,}$ so there are two in one turn → $x = \frac{\pi}{6}$

Answer: $x = \frac{3\pi}{4} \text{ or } x = \frac{5\pi}{4}$

Divide by $3\mathclose{:}$ $\cos(x) = - \frac{\sqrt{2}}{2}\mathclose{.}$ One angle is $\alpha = \arccos\left(- \frac{\sqrt{2}}{2}\right) = \frac{3\pi}{4}\mathclose{;}$ cosine is also equal at $-\alpha\mathclose{,}$ which is $2\pi - \alpha$ in one turn. So $x = \frac{3\pi}{4} \text{ or } x = \frac{5\pi}{4}\mathclose{.}$

Common mistakes, and the answer each one gives:

• Took the second solution as $\pi - \alpha\mathclose{,}$ the rule for sine → $x = \frac{\pi}{4} \text{ or } x = \frac{3\pi}{4}$
• Gave only the first solution → $x = \frac{3\pi}{4}$

Answer: $x = \frac{2\pi}{3} \text{ or } x = \frac{5\pi}{6}$

Divide by $2\mathclose{:}$ $\sin(2x) = - \frac{\sqrt{3}}{2}\mathclose{.}$ Put $u = 2x\mathclose{,}$ with $0 \le u < 2\pi\mathclose{:}$ $\sin u = - \frac{\sqrt{3}}{2}$ gives $u = \alpha$ or $u = \pi - \alpha$ with $\alpha = -\frac{\pi}{3}\mathclose{,}$ which is $\frac{5\pi}{3}$ in $[0, 2\pi)\mathclose{.}$ Halve each to get $x\mathclose{.}$ So $x = \frac{2\pi}{3} \text{ or } x = \frac{5\pi}{6}\mathclose{.}$

Common mistakes, and the answer each one gives:

• Solved for $2x$ and forgot to halve → $x = \frac{4\pi}{3} \text{ or } x = \frac{5\pi}{3}$
• Gave only the first solution → $x = \frac{2\pi}{3}$

Answer: $x = \frac{\pi}{3},\ x = \frac{2\pi}{3},\ x = \frac{4\pi}{3} \text{ or } x = \frac{5\pi}{3}$

Put $u = \cos(x)\mathclose{:}$ $4 u^{2} - 1 = 0\mathclose{,}$ so $u = \frac{1}{2}$ or $u = - \frac{1}{2}\mathclose{.}$ Every angle in one turn with $\cos(x)$ equal to $\frac{1}{2}$ or $- \frac{1}{2}\mathclose{:}$ $x = \frac{\pi}{3},\ x = \frac{2\pi}{3},\ x = \frac{4\pi}{3} \text{ or } x = \frac{5\pi}{3}\mathclose{.}$

Common mistakes, and the answer each one gives:

• Took only the first angle for each value → $x = \frac{\pi}{3} \text{ or } x = \frac{2\pi}{3}$
• Solved only the factor that gives $\cos(x) = \frac{1}{2}$ → $x = \frac{\pi}{3} \text{ or } x = \frac{5\pi}{3}$

Answer: $x = \frac{4\pi}{3} \text{ or } x = \frac{5\pi}{3}$

Divide by $6\mathclose{:}$ $\sin(x) = - \frac{\sqrt{3}}{2}\mathclose{.}$ One angle is $\alpha = \arcsin\left(- \frac{\sqrt{3}}{2}\right) = -\frac{\pi}{3}\mathclose{;}$ sine is also equal at $\pi - \alpha\mathclose{,}$ and both are brought into $[0, 2\pi)\mathclose{.}$ So $x = \frac{4\pi}{3} \text{ or } x = \frac{5\pi}{3}\mathclose{.}$

Common mistakes, and the answer each one gives:

• Took the second solution as $\pi + \alpha$ instead of $\pi - \alpha$ → $x = \frac{2\pi}{3} \text{ or } x = \frac{5\pi}{3}$
• Gave only the first solution → $x = \frac{4\pi}{3}$

Answer: $x = \frac{2\pi}{3} \text{ or } x = \frac{5\pi}{3}$

Divide by $2\mathclose{:}$ $\tan(x) = - \sqrt{3}\mathclose{.}$ One angle is $\arctan\left(- \sqrt{3}\right) = -\frac{\pi}{3}\mathclose{,}$ and tangent repeats every $\pi\mathclose{.}$ So $x = \frac{2\pi}{3} \text{ or } x = \frac{5\pi}{3}\mathclose{.}$

Common mistakes, and the answer each one gives:

• Gave one solution; tangent repeats every $\pi\mathclose{,}$ so there are two in one turn → $x = \frac{2\pi}{3}$

Answer: $x = \frac{3\pi}{4} \text{ or } x = \frac{5\pi}{4}$

Divide by $6\mathclose{:}$ $\cos(x) = - \frac{\sqrt{2}}{2}\mathclose{.}$ One angle is $\alpha = \arccos\left(- \frac{\sqrt{2}}{2}\right) = \frac{3\pi}{4}\mathclose{;}$ cosine is also equal at $-\alpha\mathclose{,}$ which is $2\pi - \alpha$ in one turn. So $x = \frac{3\pi}{4} \text{ or } x = \frac{5\pi}{4}\mathclose{.}$

Common mistakes, and the answer each one gives:

• Took the second solution as $\pi - \alpha\mathclose{,}$ the rule for sine → $x = \frac{\pi}{4} \text{ or } x = \frac{3\pi}{4}$
• Gave only the first solution → $x = \frac{3\pi}{4}$

Answer: $x = \frac{5\pi}{8} \text{ or } x = \frac{7\pi}{8}$

Divide by $3\mathclose{:}$ $\sin(2x) = - \frac{\sqrt{2}}{2}\mathclose{.}$ Put $u = 2x\mathclose{,}$ with $0 \le u < 2\pi\mathclose{:}$ $\sin u = - \frac{\sqrt{2}}{2}$ gives $u = \alpha$ or $u = \pi - \alpha$ with $\alpha = -\frac{\pi}{4}\mathclose{,}$ which is $\frac{7\pi}{4}$ in $[0, 2\pi)\mathclose{.}$ Halve each to get $x\mathclose{.}$ So $x = \frac{5\pi}{8} \text{ or } x = \frac{7\pi}{8}\mathclose{.}$

Common mistakes, and the answer each one gives:

• Solved for $2x$ and forgot to halve → $x = \frac{5\pi}{4} \text{ or } x = \frac{7\pi}{4}$
• Gave only the first solution → $x = \frac{5\pi}{8}$

Answer: $x = \frac{7\pi}{6} \text{ or } x = \frac{11\pi}{6}$

Put $u = \sin(x)\mathclose{:}$ $2 u^{2} + 5 u + 2 = 0\mathclose{,}$ so $u = -2$ or $u = - \frac{1}{2}\mathclose{.}$ A sine lies between $-1$ and $1\mathclose{,}$ so $\sin(x) = -2$ has no solution. Every angle in one turn with $\sin(x)$ equal to $- \frac{1}{2}\mathclose{:}$ $x = \frac{7\pi}{6} \text{ or } x = \frac{11\pi}{6}\mathclose{.}$

Common mistakes, and the answer each one gives:

• Took only the first angle for each value → $x = \frac{11\pi}{6}$

Answer: $x = \frac{5\pi}{4} \text{ or } x = \frac{7\pi}{4}$

Divide by $6\mathclose{:}$ $\sin(x) = - \frac{\sqrt{2}}{2}\mathclose{.}$ One angle is $\alpha = \arcsin\left(- \frac{\sqrt{2}}{2}\right) = -\frac{\pi}{4}\mathclose{;}$ sine is also equal at $\pi - \alpha\mathclose{,}$ and both are brought into $[0, 2\pi)\mathclose{.}$ So $x = \frac{5\pi}{4} \text{ or } x = \frac{7\pi}{4}\mathclose{.}$

Common mistakes, and the answer each one gives:

• Took the second solution as $\pi + \alpha$ instead of $\pi - \alpha$ → $x = \frac{3\pi}{4} \text{ or } x = \frac{7\pi}{4}$
• Gave only the first solution → $x = \frac{5\pi}{4}$

Answer: $x = \frac{\pi}{6} \text{ or } x = \frac{7\pi}{6}$

Divide by $2\mathclose{:}$ $\tan(x) = \frac{\sqrt{3}}{3}\mathclose{.}$ One angle is $\arctan\left(\frac{\sqrt{3}}{3}\right) = \frac{\pi}{6}\mathclose{,}$ and tangent repeats every $\pi\mathclose{.}$ So $x = \frac{\pi}{6} \text{ or } x = \frac{7\pi}{6}\mathclose{.}$

Common mistakes, and the answer each one gives:

• Gave one solution; tangent repeats every $\pi\mathclose{,}$ so there are two in one turn → $x = \frac{\pi}{6}$