Solving $2 \sin^2x + 6 \sin^2 \frac 12x = 3$

Question

$$ 2 \sin^2x + 6 \sin^2 \frac 12x = 3 $$ Find the angles between $0^{\circ}$ and $360^{\circ}$ which satisfy the equation above.

So, I found the answers by inputting all the possible special angles; $60^{\circ}$ and $120^{\circ}$ worked.

Are there any other ways to find the answers, and are there more than the two answers that I found?

Answer 1

HINT

Recall that

• $\sin^2 \frac x 2 =\frac12-\frac12 \cos x$
• $\sin^2x=1-\cos^2x$

therefore

$$2 \sin^2x + 6 \sin^2 \frac x 2 = 3\iff 2-2\cos^2x+3-3\cos x=3 \iff2\cos^2x+3\cos x-2=0$$

Answer 2

$2 \sin^2x + 6 \sin^2 \frac 12x = 3$

$2(1-\cos^2(x))+6\cdot\dfrac12\cdot(1-\cos(x)) = 3$

let $\cos(x) = z$

$2(1-z^2) +3(1-z) = 3$

$2-2z^2+3-3z = 3$

$2z^2+3z-2 =0$

Solving the above quadratic and substituting back $z =\cos(x)$ will give you the solutions.