Trigonometric equations

Example 1: Solve ${\sin x=\dfrac{1}{2}}$.

Solution:

Step 1: Solve in one period. In ${[0,2\pi)}$, sine is positive in quadrants I and II. The reference angle is ${\dfrac{\pi}{6}}$. So

${x=\dfrac{\pi}{6}}$or${x=\pi-\dfrac{\pi}{6}=\dfrac{5\pi}{6}}$

Step 2: Add multiples of the period. Sine has period ${2\pi}$, so every solution is one of these plus a whole number of full turns:

${x=\dfrac{\pi}{6}+2k\pi}$or${x=\dfrac{5\pi}{6}+2k\pi}$

Here ${k}$ is any integer (${0}$, ${\pm 1}$, ${\pm 2}$, and so on).

Example 2: Solve ${2\cos x+\sqrt{3}=0}$.

Solution:

Get cosine alone:

${\cos x=-\dfrac{\sqrt{3}}{2}}$

Cosine is negative in quadrants II and III. The reference angle is ${\dfrac{\pi}{6}}$. So in ${[0,2\pi)}$:

${x=\pi-\dfrac{\pi}{6}=\dfrac{5\pi}{6}}$or${x=\pi+\dfrac{\pi}{6}=\dfrac{7\pi}{6}}$

All solutions:

${x=\dfrac{5\pi}{6}+2k\pi}$or${x=\dfrac{7\pi}{6}+2k\pi}$

Example 3: Solve ${\tan x=-1}$.

Solution:

Tangent has period ${\pi}$, so find the solutions in one interval of length ${\pi}$, such as $\left(-\dfrac{\pi}{2},\dfrac{\pi}{2}\right)$. There, ${\tan x=-1}$ only at ${x=-\dfrac{\pi}{4}}$. All solutions:

${x=-\dfrac{\pi}{4}+k\pi}$

In ${[0,2\pi)}$, these are ${\dfrac{3\pi}{4}}$ (with ${k=1}$) and ${\dfrac{7\pi}{4}}$ (with ${k=2}$).

Example 4: Find the solutions of ${\sin x=0.3}$ in ${[0,2\pi)}$. Round to four decimal places.

Solution:

This is not a special value, so use a calculator in radian mode:

${\sin^{-1}(0.3)\approx 0.3047}$

That is the solution in quadrant I. Sine is also positive in quadrant II, at ${\pi}$ minus the reference angle:

${x\approx\pi-0.3047\approx 2.8369}$

The solutions are ${x\approx 0.3047}$ and ${x\approx 2.8369}$.

Example 5: Find the solutions of ${\sin 2x=\dfrac{\sqrt{3}}{2}}$ in ${[0,2\pi)}$.

Solution:

Let ${u=2x}$. Then ${\sin u=\dfrac{\sqrt{3}}{2}}$, so

${u=\dfrac{\pi}{3}+2k\pi}$or${u=\dfrac{2\pi}{3}+2k\pi}$

Now ${x=\dfrac{u}{2}}$. Divide everything by ${2}$:

${x=\dfrac{\pi}{6}+k\pi}$or${x=\dfrac{\pi}{3}+k\pi}$

Take ${k=0}$ and ${k=1}$ to stay in ${[0,2\pi)}$:

${x=\dfrac{\pi}{6}}$, ${\dfrac{\pi}{3}}$, ${\dfrac{7\pi}{6}}$, ${\dfrac{4\pi}{3}}$

Example 6: Find the solutions of ${2\sin^2 x-\sin x-1=0}$ in ${[0,2\pi)}$.

Solution:

This is like ${2s^2-s-1=0}$ with ${s=\sin x}$. It factors as ${(2s+1)(s-1)=0}$:

${(2\sin x+1)(\sin x-1)=0}$

So ${\sin x=-\dfrac{1}{2}}$ or ${\sin x=1}$.

• ${\sin x=-\dfrac{1}{2}}$: sine is negative in quadrants III and IV, reference angle ${\dfrac{\pi}{6}}$. So ${x=\dfrac{7\pi}{6}}$ or ${x=\dfrac{11\pi}{6}}$.
• ${\sin x=1}$: ${x=\dfrac{\pi}{2}}$.

The solutions are ${\dfrac{\pi}{2}}$, ${\dfrac{7\pi}{6}}$, and ${\dfrac{11\pi}{6}}$.

Example 7: Find the solutions of ${\tan x\sin x=\sin x}$ in ${[0,2\pi)}$.

Solution:

Move everything to one side, and factor out ${\sin x}$:

$\begin{align*}\tan x\sin x-\sin x&=0\\\sin x\,(\tan x-1)&=0\end{align*}$

So ${\sin x=0}$ or ${\tan x=1}$.

• ${\sin x=0}$: ${x=0}$ or ${x=\pi}$.
• ${\tan x=1}$: ${x=\dfrac{\pi}{4}}$ or ${x=\dfrac{5\pi}{4}}$.

The solutions are ${0}$, ${\dfrac{\pi}{4}}$, ${\pi}$, and ${\dfrac{5\pi}{4}}$.