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

The general solution for 2sin^2(x)=cos(x)+1 is x=pi+2pin,x= pi/3+2pin,x=(5pi)/3+2pin

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

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

x = π + 2 πn, x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

Solution steps

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

Subtract

cos (x) + 1

from both sides

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

Rewrite using trig identities

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

Solve by substitution

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

cos (x) = - 1 : x = π + 2 πn

cos (x) = \frac{1}{2} : x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

Combine all the solutions

x = π + 2 πn, x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

2 cos^{2} (θ) + cos (θ) = 0

9 sec (x) + 9 tan (x) = 9

sin (2 θ) + 1 = 0

x, cos (x) = cos (y)

3 tanh^{2} (θ) = 5 sech (θ) + 1

What is the general solution for 2sin^2(x)=cos(x)+1 ?
The general solution for 2sin^2(x)=cos(x)+1 is x=pi+2pin,x= pi/3+2pin,x=(5pi)/3+2pin