What is the general solution for 4cos(2θ)=8sin^2(θ) ?

The general solution for 4cos(2θ)=8sin^2(θ) is θ=(5pi)/6+pin,θ= pi/6+pin

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4cos(2θ)=8sin^2(θ)

4 cos (2 θ) = 8 sin^{2} (θ)

θ = \frac{5 π}{6} + πn, θ = \frac{π}{6} + πn

Solution steps

4 cos (2 θ) = 8 sin^{2} (θ)

Subtract

8 sin^{2} (θ)

from both sides

4 cos (2 θ) - 8 sin^{2} (θ) = 0

Rewrite using trig identities

- 12 sin^{2} (θ) + 4 cos^{2} (θ) = 0

Factor

- 12 sin^{2} (θ) + 4 cos^{2} (θ) : 4 (cos (θ) + 3 sin (θ)) (cos (θ) - 3 sin (θ))

4 (cos (θ) + 3 sin (θ)) (cos (θ) - 3 sin (θ)) = 0

Solving each part separately

cos (θ) + 3 sin (θ) = 0 or cos (θ) - 3 sin (θ) = 0

cos (θ) + 3 sin (θ) = 0 : θ = \frac{5 π}{6} + πn

cos (θ) - 3 sin (θ) = 0 : θ = \frac{π}{6} + πn

Combine all the solutions

θ = \frac{5 π}{6} + πn, θ = \frac{π}{6} + πn

cos (θ) = (\frac{5}{29})

sin (θ) = \frac{7.5}{6.5}

cos^{2} (x) - \frac{3}{2} = \frac{5}{2} cos (x)

θ, w = fd cos (θ)

sec (x) - sec (x) tan (x) = 0

What is the general solution for 4cos(2θ)=8sin^2(θ) ?
The general solution for 4cos(2θ)=8sin^2(θ) is θ=(5pi)/6+pin,θ= pi/6+pin