Solve each equation for 0

Question

Solve each equation for 0<x<2pi:
(2sinx-1)(2cos^2x-1)=0

anonymous · Answer

please i would like an explanation not a reference

anonymous · Answer

(2sinx-1) = 0 
or 
(2cos^2x-1)=0

anonymous · Answer

the product of both is 0

anonymous · Answer

for that to happen, one or the other must equal 0. like with quadratics

anonymous · Answer

$$\sin(x) = 1/2$$or$$\cos(x) = \sqrt{2}/2$$

anonymous · Answer

do you understand?

anonymous · Answer

you just set each one equal to zero...however im using the sum and difference identity so im confused as to how can i use both the cosine and sine you found to find specific degrees

anonymous · Answer

why would you use the sum and difference formula? its like a factored quadratic

anonymous · Answer

oops, my mistake. i was thinking of another problem. but for this problem, can only one be equal to 0? could you write out step by step how you would find the final solution?

anonymous · Answer

afteri get sin(x)=1/2 how would the whole equation look since the cosine part of the equation wasnt solved for?