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

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

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8-sin(θ)=cos(2θ)

8 - sin (θ) = cos (2 θ)

No Solution for θ \in R

Solution steps

8 - sin (θ) = cos (2 θ)

Subtract

cos (2 θ)

from both sides

8 - sin (θ) - cos (2 θ) = 0

Rewrite using trig identities

7 - sin (θ) + 2 sin^{2} (θ) = 0

Solve by substitution

sin (θ) = \frac{1}{4} + i \frac{55}{4}, sin (θ) = \frac{1}{4} - i \frac{55}{4}

sin (θ) = \frac{1}{4} + i \frac{55}{4} :

No Solution

sin (θ) = \frac{1}{4} - i \frac{55}{4} :

No Solution

Combine all the solutions

No Solution for θ \in R

x, 2 cos^{2} (x) - cos (x) - 1 = 0

- 8 cos (- 4 θ + 150) = 43

4 cos^{2} (x) + 2 sin (x) - 1 = 0

cos (x) = \frac{- 3}{2}, 0 \leq x \leq 2 π

- sin (t) = cos (t)

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