What is the general solution for 3sin^2(θ)+2cos(2θ)=1,0<= θ<= 2pi ?

The general solution for 3sin^2(θ)+2cos(2θ)=1,0<= θ<= 2pi is θ= pi/2 ,θ=(3pi)/2

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3sin^2(θ)+2cos(2θ)=1,0<= θ<= 2pi

3 sin^{2} (θ) + 2 cos (2 θ) = 1, 0 \leq θ \leq 2 π

θ = \frac{π}{2}, θ = \frac{3 π}{2}

Solution steps

3 sin^{2} (θ) + 2 cos (2 θ) = 1, 0 \leq θ \leq 2 π

Subtract

1

from both sides

3 sin^{2} (θ) + 2 cos (2 θ) - 1 = 0

Rewrite using trig identities

1 - sin^{2} (θ) = 0

Solve by substitution

sin (θ) = 1, sin (θ) = - 1

sin (θ) = 1, 0 \leq θ \leq 2 π : θ = \frac{π}{2}

sin (θ) = - 1, 0 \leq θ \leq 2 π : θ = \frac{3 π}{2}

Combine all the solutions

θ = \frac{π}{2}, θ = \frac{3 π}{2}

2 cos^{2} (x) + 3 sin (- x) - 3 = 0

12 cos (\frac{x}{3}) = 0

x, y^{4} + 2 y + 1 = sin (x)

cos (2 x) = \frac{( 6 k - 9 )}{9}

csc (x) = 1 + cot (x)

What is the general solution for 3sin^2(θ)+2cos(2θ)=1,0<= θ<= 2pi ?
The general solution for 3sin^2(θ)+2cos(2θ)=1,0<= θ<= 2pi is θ= pi/2 ,θ=(3pi)/2