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]?

1 Answer
Oct 26, 2014

2sin^2(x/4)-3cos(x/4)=0

by the trig identity sin^2theta=1-cos^2theta,

Rightarrow 2-2cos^2(x/4)-3cos(x/4)=0

by multiplying by (-1) and rearranging terms,

Rightarrow 2cos^2(x/4)+3cos(x/4)-2=0

by factoring out,

Rightarrow [2cos(x/4)-1][cos(x/4)+2]=0

since cos(x/4)+2 ne0,

Rightarrow2cos(x/4)-1=0

Rightarrow cos(x/4)=1/2 Rightarrow x/4=pi/6Rightarrow x={4pi}/6={2pi}/3
(Note: There are more x-values satisfying the equation, but other ones are not in [0, 2pi].)


I hope that this was helpful.