What is the general solution for cos(2θ)+14sin^2(θ)=10,0<θ<2pi ?

The general solution for cos(2θ)+14sin^2(θ)=10,0<θ<2pi is θ= pi/3 ,θ=(2pi)/3 ,θ=(4pi)/3 ,θ=(5pi)/3

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cos(2θ)+14sin^2(θ)=10,0<θ<2pi

cos (2 θ) + 14 sin^{2} (θ) = 10, 0 < 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) + 14 sin^{2} (θ) = 10, 0 < θ < 2 π

Subtract

10

from both sides

cos (2 θ) + 14 sin^{2} (θ) - 10 = 0

Rewrite using trig identities

- 9 + 12 sin^{2} (θ) = 0

Solve by substitution

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

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

sin (θ) = - \frac{3}{2}, 0 < 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}

14 sin^{2} (x) + 5 sin (x) - 1 = 0

sin (x) = \frac{3}{5} sin (x + \frac{π}{2}) + cos (π - x)

cos (x + π) - sin (x - π) = 0

2 sin (x) tan (x) - tan (x) - 2 sin (x) + 1 = 0

sin (2 x) = \frac{( 6 m - 5 )}{5}

What is the general solution for cos(2θ)+14sin^2(θ)=10,0<θ<2pi ?
The general solution for cos(2θ)+14sin^2(θ)=10,0<θ<2pi is θ= pi/3 ,θ=(2pi)/3 ,θ=(4pi)/3 ,θ=(5pi)/3