Question

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]`?

Answer 1

`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]`.)

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I hope that this was helpful.