How do you find all the solutions for 2 \sin^2 \frac{x}{4}-3 \cos \frac{x}{4} = 0 over the interval [0,2pi]?

Oct 26, 2014

$2 {\sin}^{2} \left(\frac{x}{4}\right) - 3 \cos \left(\frac{x}{4}\right) = 0$

by the trig identity ${\sin}^{2} \theta = 1 - {\cos}^{2} \theta$,

$R i g h t a r r o w 2 - 2 {\cos}^{2} \left(\frac{x}{4}\right) - 3 \cos \left(\frac{x}{4}\right) = 0$

by multiplying by $\left(- 1\right)$ and rearranging terms,

$R i g h t a r r o w 2 {\cos}^{2} \left(\frac{x}{4}\right) + 3 \cos \left(\frac{x}{4}\right) - 2 = 0$

by factoring out,

$R i g h t a r r o w \left[2 \cos \left(\frac{x}{4}\right) - 1\right] \left[\cos \left(\frac{x}{4}\right) + 2\right] = 0$

since $\cos \left(\frac{x}{4}\right) + 2 \ne 0$,

$R i g h t a r r o w 2 \cos \left(\frac{x}{4}\right) - 1 = 0$

$R i g h t a r r o w \cos \left(\frac{x}{4}\right) = \frac{1}{2} R i g h t a r r o w \frac{x}{4} = \frac{\pi}{6} R i g h t a r r o w x = \frac{4 \pi}{6} = \frac{2 \pi}{3}$
(Note: There are more $x$-values satisfying the equation, but other ones are not in $\left[0 , 2 \pi\right]$.)

I hope that this was helpful.