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

The general solution for 2cos^2(2θ)+sin(2θ)=1,-180<= θ<= 180 is θ=105,θ=165,θ=-75,θ=-15,θ=45,θ=-135

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2cos^2(2θ)+sin(2θ)=1,-180<= θ<= 180

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

θ = 10 5^{\circ}, θ = 16 5^{\circ}, θ = - 7 5^{\circ}, θ = - 1 5^{\circ}, θ = 4 5^{\circ}, θ = - 13 5^{\circ}

Solution steps

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

Subtract

1

from both sides

2 cos^{2} (2 θ) + sin (2 θ) - 1 = 0

Rewrite using trig identities

1 + sin (2 θ) - 2 sin^{2} (2 θ) = 0

Solve by substitution

sin (2 θ) = - \frac{1}{2}, sin (2 θ) = 1

sin (2 θ) = - \frac{1}{2}, - 18 0^{\circ} \leq θ \leq 18 0^{\circ} : θ = 10 5^{\circ}, θ = 16 5^{\circ}, θ = - 7 5^{\circ}, θ = - 1 5^{\circ}

sin (2 θ) = 1, - 18 0^{\circ} \leq θ \leq 18 0^{\circ} : θ = 4 5^{\circ}, θ = - 13 5^{\circ}

Combine all the solutions

θ = 10 5^{\circ}, θ = 16 5^{\circ}, θ = - 7 5^{\circ}, θ = - 1 5^{\circ}, θ = 4 5^{\circ}, θ = - 13 5^{\circ}

3 cos (4 x) - 2 = \frac{3}{2} sin (8 x) - 2

4 \cdot cos (x) sin (x) + 3 \cdot cos (x) = 0

(cot (x)) (tan (x) - 1) = 0

10 tan (θ) sec (θ) = 10 cot (θ) csc (θ)

arctan (x) = \frac{( 2 \cdot 1.8 )}{[ ( 3170 + 1.7 ) - 1 ]}

What is the general solution for 2cos^2(2θ)+sin(2θ)=1,-180<= θ<= 180 ?
The general solution for 2cos^2(2θ)+sin(2θ)=1,-180<= θ<= 180 is θ=105,θ=165,θ=-75,θ=-15,θ=45,θ=-135