What is the general solution for cos(2θ)(4sin^2(θ)+1)=0 ?

The general solution for cos(2θ)(4sin^2(θ)+1)=0 is θ= pi/4+pin,θ=(3pi)/4+pin

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cos(2θ)(4sin^2(θ)+1)=0

cos (2 θ) (4 sin^{2} (θ) + 1) = 0

θ = \frac{π}{4} + πn, θ = \frac{3 π}{4} + πn

Solution steps

cos (2 θ) (4 sin^{2} (θ) + 1) = 0

Solving each part separately

cos (2 θ) = 0 or 4 sin^{2} (θ) + 1 = 0

cos (2 θ) = 0 : θ = \frac{π}{4} + πn, θ = \frac{3 π}{4} + πn

4 sin^{2} (θ) + 1 = 0 :

No Solution

Combine all the solutions

θ = \frac{π}{4} + πn, θ = \frac{3 π}{4} + πn

sin (x) = 0.9906

sin (θ) = - 0.17

2 sin^{2} (x) - 1 = 0, - π \leq x \leq π

sin (θ) = - 0.22

\frac{3}{2} sec (θ) + 3 cos (θ) = 3

What is the general solution for cos(2θ)(4sin^2(θ)+1)=0 ?
The general solution for cos(2θ)(4sin^2(θ)+1)=0 is θ= pi/4+pin,θ=(3pi)/4+pin