What is the general solution for cos(2θ)+6sin^2(θ)=4,0<= θ<2pi ?

The general solution for cos(2θ)+6sin^2(θ)=4,0<= θ<2pi is θ= pi/3 ,θ=(2pi)/3 ,θ=(4pi)/3 ,θ=(5pi)/3

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cos(2θ)+6sin^2(θ)=4,0<= θ<2pi

cos (2 θ) + 6 sin^{2} (θ) = 4, 0 \leq θ < 2 π

θ = \frac{π}{3}, θ = \frac{2 π}{3}, θ = \frac{4 π}{3}, θ = \frac{5 π}{3}

Solution steps

cos (2 θ) + 6 sin^{2} (θ) = 4, 0 \leq θ < 2 π

Subtract

4

from both sides

cos (2 θ) + 6 sin^{2} (θ) - 4 = 0

Rewrite using trig identities

- 3 + 4 sin^{2} (θ) = 0

Solve by substitution

sin (θ) = \frac{3}{2}, sin (θ) = - \frac{3}{2}

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

sin (θ) = - \frac{3}{2}, 0 \leq θ < 2 π : θ = \frac{4 π}{3}, θ = \frac{5 π}{3}

Combine all the solutions

θ = \frac{π}{3}, θ = \frac{2 π}{3}, θ = \frac{4 π}{3}, θ = \frac{5 π}{3}

tan (x) = \frac{6}{9}

3 csc (θ) - 2 = 0

2 tan (2 x) + 2 = 0

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

sin (α) - cos (α) = m

What is the general solution for cos(2θ)+6sin^2(θ)=4,0<= θ<2pi ?
The general solution for cos(2θ)+6sin^2(θ)=4,0<= θ<2pi is θ= pi/3 ,θ=(2pi)/3 ,θ=(4pi)/3 ,θ=(5pi)/3