What is the general solution for cos(2θ)=-sin^2(θ) ?

The general solution for cos(2θ)=-sin^2(θ) is θ= pi/2+2pin,θ=(3pi)/2+2pin

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cos(2θ)=-sin^2(θ)

cos (2 θ) = - sin^{2} (θ)

θ = \frac{π}{2} + 2 πn, θ = \frac{3 π}{2} + 2 πn

Solution steps

cos (2 θ) = - sin^{2} (θ)

Subtract

- sin^{2} (θ)

from both sides

cos (2 θ) + sin^{2} (θ) = 0

Rewrite using trig identities

cos^{2} (θ) = 0

Apply rule

x^{n} = 0 \Rightarrow x = 0

cos (θ) = 0

General solutions for

cos (θ) = 0

θ = \frac{π}{2} + 2 πn, θ = \frac{3 π}{2} + 2 πn

10 cos (x) = 0

3 sin (x) - 2 = - 2

2 sin (x) cos^{2} (x) = - \frac{3}{2} cos (x)

4 sin (x^{\circ}) + 1 = - 2

(2 cos (x) + 1) (3 tan (x) - 1) = 0

What is the general solution for cos(2θ)=-sin^2(θ) ?
The general solution for cos(2θ)=-sin^2(θ) is θ= pi/2+2pin,θ=(3pi)/2+2pin