Question

How do you solve `2(sin2x)^2 + 5sin2x - 3 = 0`?

Answer 1

The value of `x=15^o or x=75^o`

Explanation:

We set `t=sin2x` so we get the trinomial

`2t^2+5t-3=0`

Which has roots `r_1=-3,r_2=1/2` .Only `r_2=1/2` is acceptable

hence we have that

`sin2x=1/2=>sin2x=sin30 or sin2x=sin150=>x=15^o,75^o`