What is the general solution for -2cos(2θ)+12=4sin(θ)+9 ?

The general solution for -2cos(2θ)+12=4sin(θ)+9 is θ= pi/6+2pin,θ=(5pi)/6+2pin

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-2cos(2θ)+12=4sin(θ)+9

- 2 cos (2 θ) + 12 = 4 sin (θ) + 9

θ = \frac{π}{6} + 2 πn, θ = \frac{5 π}{6} + 2 πn

Solution steps

- 2 cos (2 θ) + 12 = 4 sin (θ) + 9

Subtract

4 sin (θ) + 9

from both sides

- 2 cos (2 θ) - 4 sin (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{π}{6} + 2 πn, θ = \frac{5 π}{6} + 2 πn

Combine all the solutions

θ = \frac{π}{6} + 2 πn, θ = \frac{5 π}{6} + 2 πn

sec^{2} (x) + sec (x) - 2 = 0

(cot (x) - 1) (2 sin (x) + 1) = 0

y, x = tan (y)

tan (x) = - \frac{12}{5}

2 cos (x) = sin (x)

What is the general solution for -2cos(2θ)+12=4sin(θ)+9 ?
The general solution for -2cos(2θ)+12=4sin(θ)+9 is θ= pi/6+2pin,θ=(5pi)/6+2pin