Trig mk4

Again with the last thread 0°≤x≤360°

4(sin²x-cosx)=3-2cosx

Here's what I've done

4(1-cos²x-cosx)=3-2cosx
4-4cos²x-4cosx=3-2cosx
1-4cos²x-2cosx=0
4cos²x+2cosx=1
2cosx(2cosx+1)=1


2cosx=1
where cosx=0.5
x= 60 and 300 (Using CAST diagram)

2cosx+1=1
where 2cosx=0
x=90 and 270 (using graph of cos)

where did i go wrong?

2cosx(2cosx+1)=1


2cosx=1

You can only do that if it's =0, you have =1, make it a quadratic in cos x =0 and then solve.

lol just realised xD

no worries ...
(hopefully you liked the picture ...)