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

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

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2sin^2(x)=2+cos(x),0<= x<= 2pi

2 sin^{2} (x) = 2 + cos (x), 0 \leq x \leq 2 π

x = \frac{2 π}{3}, x = \frac{4 π}{3}, x = \frac{π}{2}, x = \frac{3 π}{2}

Solution steps

2 sin^{2} (x) = 2 + cos (x), 0 \leq x \leq 2 π

Subtract

2 + cos (x)

from both sides

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

Rewrite using trig identities

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

Solve by substitution

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

cos (x) = - \frac{1}{2}, 0 \leq x \leq 2 π : x = \frac{2 π}{3}, x = \frac{4 π}{3}

cos (x) = 0, 0 \leq x \leq 2 π : x = \frac{π}{2}, x = \frac{3 π}{2}

Combine all the solutions

x = \frac{2 π}{3}, x = \frac{4 π}{3}, x = \frac{π}{2}, x = \frac{3 π}{2}

2 sin (θ) tan (θ) - tan (θ) = 0

sec (x) + 2 = 0

sin (x) + 1 = 0, 0 \leq x \leq 2 π

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

12 tan (x) + 5 = 5 tan (x) + 5

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