Question

Solve algebraically for 0≤x≤2π, giving answers in terms of π:
a. sin^2 x-sinx-2=0
b. 4sin^2 x=1

Answer 1

a. This is equation of quadratic type, substitute u=sinx and you get \[u^2-u-2=0\] which conveniently factors \[(u-2)(u+1)=0\]so you get sinx=2 or sinx=1. Since 2 is not in the range of sinx, that one is extraneous. For sinx=1, there is only one real number 0≤x≤2π such that sinx=1 .... what is it? That's your solution.

b. \[4\sin ^{2}x=1\] \[\sin ^{2}x=\frac{1}{4}\]\[\sin x=\pm\sqrt{\frac{1}{4}}\]\[\sin x=\pm\frac{1}{2}\]So how many solutions do you get for 0≤x≤2π?