What is the general solution for 2cos^2(θ)-1=sin(θ),0<= θ<= 180 ?

The general solution for 2cos^2(θ)-1=sin(θ),0<= θ<= 180 is θ=30,θ=150

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2cos^2(θ)-1=sin(θ),0<= θ<= 180

2 cos^{2} (θ) - 1 = sin (θ), 0^{\circ} \leq θ \leq 18 0^{\circ}

θ = 3 0^{\circ}, θ = 15 0^{\circ}

Solution steps

2 cos^{2} (θ) - 1 = sin (θ), 0^{\circ} \leq θ \leq 18 0^{\circ}

Subtract

sin (θ)

from both sides

2 cos^{2} (θ) - 1 - sin (θ) = 0

Rewrite using trig identities

1 - sin (θ) - 2 sin^{2} (θ) = 0

Solve by substitution

sin (θ) = - 1, sin (θ) = \frac{1}{2}

sin (θ) = - 1, 0 \leq θ \leq 18 0^{\circ} :

No Solution

sin (θ) = \frac{1}{2}, 0 \leq θ \leq 18 0^{\circ} : θ = 3 0^{\circ}, θ = 15 0^{\circ}

Combine all the solutions

θ = 3 0^{\circ}, θ = 15 0^{\circ}

x, e^{c o s (x)} (1 - cos (y)) = c

22 = 2 sec (θ)

cos^{4} (x) - sin^{4} (x) = \frac{1}{2}

sin (x) + cos (x) = tan (x) + 1

cos (x) = \frac{1}{4}, 0 \leq x \leq 2 π

What is the general solution for 2cos^2(θ)-1=sin(θ),0<= θ<= 180 ?
The general solution for 2cos^2(θ)-1=sin(θ),0<= θ<= 180 is θ=30,θ=150