Question

Solve the equation on the interval 0 ≤ θ < 2π.

2 sin^2 θ = 3(cos θ + 1)

Answer 1

So far I got

Sin^2x = 3(cosx+1) /2

Answer 2

Can Someone give me some basics about this because I feel lost

Answer 3

\[\textbf{HINT}\]\[\cos^2\theta = {\cos\theta + 1 \over 2}\]

Answer 4

So you have\[\cos ^2 \theta = 3\sin^2 \theta\]

Answer 5

No, I will have\[\sin ^{2} = 3 \cos ^{2} x\]

Answer 6

But the variable in your question is \(\theta\) :P