What is the general solution for -2cos(2θ)-5=-4sin(θ)-8 ?

The general solution for -2cos(2θ)-5=-4sin(θ)-8 is θ=(7pi)/6+2pin,θ=(11pi)/6+2pin

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-2cos(2θ)-5=-4sin(θ)-8

- 2 cos (2 θ) - 5 = - 4 sin (θ) - 8

θ = \frac{7 π}{6} + 2 πn, θ = \frac{11 π}{6} + 2 πn

Solution steps

- 2 cos (2 θ) - 5 = - 4 sin (θ) - 8

Subtract

- 4 sin (θ) - 8

from both sides

4 sin (θ) - 2 cos (2 θ) + 3 = 0

Rewrite using trig identities

1 + 4 sin (θ) + 4 sin^{2} (θ) = 0

Solve by substitution

sin (θ) = - \frac{1}{2}

sin (θ) = - \frac{1}{2} : θ = \frac{7 π}{6} + 2 πn, θ = \frac{11 π}{6} + 2 πn

Combine all the solutions

θ = \frac{7 π}{6} + 2 πn, θ = \frac{11 π}{6} + 2 πn

3 cos^{2} (θ) - cos (θ) = 0

10 sin^{2} (x) + 11 sin (x) - 8 = 0

2 cos^{2} (x) + 7 cos (x) - 4 = 0

cot (2 θ) = \frac{3}{3}

4 sin^{2} (\frac{θ}{2}) + 8 cos (\frac{θ}{2}) - 7 = 0, 0 \leq θ \leq 36 0^{\circ}

What is the general solution for -2cos(2θ)-5=-4sin(θ)-8 ?
The general solution for -2cos(2θ)-5=-4sin(θ)-8 is θ=(7pi)/6+2pin,θ=(11pi)/6+2pin