Skill: Solve trigonometric equations over specified intervals

To solve a trigonometric equation on a specified interval:

1. Rewrite or factor the equation until you can identify basic trigonometric values.
2. Find the reference angle using an inverse trigonometric function.
3. Use the signs in each quadrant to find all possible angles.
4. Keep only angles in the given interval.
5. Check the solutions in the original equation.

Example

Solve

$$2\sin^2 x-3\sin x+1=0$$

for

$$0^\circ\le x\le 360^\circ.$$

Step 1: Factor the equation

Treat $\sin x$ like a variable. The expression factors as

$$2\sin^2 x-3\sin x+1=(2\sin x-1)(\sin x-1).$$

Therefore,

$$(2\sin x-1)(\sin x-1)=0.$$

Using the zero-product property, either

$$2\sin x-1=0$$

or

$$\sin x-1=0.$$

So we solve the two equations:

$$\sin x=\frac12$$

and

$$\sin x=1.$$

Step 2: Solve $\sin x=\frac12$

The reference angle is

$$x_r=\sin^{-1}\left(\frac12\right)=30^\circ.$$

Sine is positive in Quadrants I and II. Therefore, the angles are

$$x=30^\circ$$

and

$$x=180^\circ-30^\circ=150^\circ.$$

Step 3: Solve $\sin x=1$

Sine equals $1$ at

$$x=90^\circ.$$

Step 4: List the solutions in the interval

All three angles lie between $0^\circ$ and $360^\circ$, so the solutions are

$$\boxed{x=30^\circ,\ 90^\circ,\ 150^\circ}.$$

Step 5: Check

For $x=30^\circ$ and $x=150^\circ$, $\sin x=\frac12$:

$$2\left(\frac12\right)^2-3\left(\frac12\right)+1 =\frac12-\frac32+1=0.$$

For $x=90^\circ$, $\sin x=1$:

$$2(1)^2-3(1)+1=2-3+1=0.$$

All three values satisfy the original equation.