What is the general solution for 2cos^2(x)=1-sin(x) ?

The general solution for 2cos^2(x)=1-sin(x) is x=(7pi)/6+2pin,x=(11pi)/6+2pin,x= pi/2+2pin

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2cos^2(x)=1-sin(x)

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

x = \frac{7 π}{6} + 2 πn, x = \frac{11 π}{6} + 2 πn, x = \frac{π}{2} + 2 πn

Solution steps

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

Subtract

1 - sin (x)

from both sides

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

Rewrite using trig identities

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

Solve by substitution

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

sin (x) = - \frac{1}{2} : x = \frac{7 π}{6} + 2 πn, x = \frac{11 π}{6} + 2 πn

sin (x) = 1 : x = \frac{π}{2} + 2 πn

Combine all the solutions

x = \frac{7 π}{6} + 2 πn, x = \frac{11 π}{6} + 2 πn, x = \frac{π}{2} + 2 πn

sin (t) = - \frac{3}{2}

2 sin (2 θ) - 3 sin (θ) = 0

cos^{2} (x) + cos^{2} (\frac{x}{2}) - \frac{1}{2} = 0

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

cos (2 x) + 3 sin (x) = 1

What is the general solution for 2cos^2(x)=1-sin(x) ?
The general solution for 2cos^2(x)=1-sin(x) is x=(7pi)/6+2pin,x=(11pi)/6+2pin,x= pi/2+2pin