What is the general solution for 2sin^2(x)-cos(x)=1,0<= x<= 2pi ?

The general solution for 2sin^2(x)-cos(x)=1,0<= x<= 2pi is x=pi,x= pi/3 ,x=(5pi)/3

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2sin^2(x)-cos(x)=1,0<= x<= 2pi

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

x = π, x = \frac{π}{3}, x = \frac{5 π}{3}

Solution steps

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

Subtract

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, 0 \leq x \leq 2 π : x = π

cos (x) = \frac{1}{2}, 0 \leq x \leq 2 π : x = \frac{π}{3}, x = \frac{5 π}{3}

Combine all the solutions

x = π, x = \frac{π}{3}, x = \frac{5 π}{3}

cos (θ) = - \frac{π}{2}

4 sin (t^{2}) - 1 = 1

sec (x) + 4 = 4 sec (x)

cos (x) + 2 cos (2 x) = 0

tan (θ) = - 15

What is the general solution for 2sin^2(x)-cos(x)=1,0<= x<= 2pi ?
The general solution for 2sin^2(x)-cos(x)=1,0<= x<= 2pi is x=pi,x= pi/3 ,x=(5pi)/3