What is the general solution for -5cos(2θ)-9sin(θ)+8=-sin(θ) ?

The general solution for -5cos(2θ)-9sin(θ)+8=-sin(θ) is No Solution for θ\in\mathbb{R}

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-5cos(2θ)-9sin(θ)+8=-sin(θ)

- 5 cos (2 θ) - 9 sin (θ) + 8 = - sin (θ)

No Solution for θ \in R

Solution steps

- 5 cos (2 θ) - 9 sin (θ) + 8 = - sin (θ)

Subtract

- sin (θ)

from both sides

- 5 cos (2 θ) - 8 sin (θ) + 8 = 0

Rewrite using trig identities

3 + 10 sin^{2} (θ) - 8 sin (θ) = 0

Solve by substitution

sin (θ) = \frac{2}{5} + i \frac{14}{10}, sin (θ) = \frac{2}{5} - i \frac{14}{10}

sin (θ) = \frac{2}{5} + i \frac{14}{10} :

No Solution

sin (θ) = \frac{2}{5} - i \frac{14}{10} :

No Solution

Combine all the solutions

No Solution for θ \in R

3 + 3 cos (θ) = 3 + 3 sin (θ)

sinh (x) = \frac{4}{3}

x, f^{- 1} = arctan (x^{3} - 1)

3 cos (B) + 3 = 0

sin (2 θ) = \frac{3}{4}

What is the general solution for -5cos(2θ)-9sin(θ)+8=-sin(θ) ?
The general solution for -5cos(2θ)-9sin(θ)+8=-sin(θ) is No Solution for θ\in\mathbb{R}