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

The general solution for 2cos^2(2θ)+sin(2θ)=1 is θ=(7pi)/(12)+pin,θ=(11pi)/(12)+pin,θ= pi/4+pin

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2cos^2(2θ)+sin(2θ)=1

2 cos^{2} (2 θ) + sin (2 θ) = 1

θ = \frac{7 π}{12} + πn, θ = \frac{11 π}{12} + πn, θ = \frac{π}{4} + πn

Solution steps

2 cos^{2} (2 θ) + sin (2 θ) = 1

Subtract

1

from both sides

2 cos^{2} (2 θ) + sin (2 θ) - 1 = 0

Rewrite using trig identities

1 + sin (2 θ) - 2 sin^{2} (2 θ) = 0

Solve by substitution

sin (2 θ) = - \frac{1}{2}, sin (2 θ) = 1

sin (2 θ) = - \frac{1}{2} : θ = \frac{7 π}{12} + πn, θ = \frac{11 π}{12} + πn

sin (2 θ) = 1 : θ = \frac{π}{4} + πn

Combine all the solutions

θ = \frac{7 π}{12} + πn, θ = \frac{11 π}{12} + πn, θ = \frac{π}{4} + πn

8 tan (θ) - 54 = 0

3 sin^{2} (x) - 3 = 0

tan (2 x + \frac{π}{3}) = 3

tan (θ) = - 8

2 cos (2 θ) + 7 sin (θ) - 2 = - 2

What is the general solution for 2cos^2(2θ)+sin(2θ)=1 ?
The general solution for 2cos^2(2θ)+sin(2θ)=1 is θ=(7pi)/(12)+pin,θ=(11pi)/(12)+pin,θ= pi/4+pin