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

The general solution for 2cos^2(x)+sin(x)-1=0,0<= x<= 2pi is x=(7pi)/6 ,x=(11pi)/6 ,x= pi/2

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

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

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

Solution steps

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

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}, 0 \leq x \leq 2 π : x = \frac{7 π}{6}, x = \frac{11 π}{6}

sin (x) = 1, 0 \leq x \leq 2 π : x = \frac{π}{2}

Combine all the solutions

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

cos^{3} (x) = cos^{2} (x)

cos (θ) = - 0.8

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

tan^{2} (x) + 3 tan (x) = - 1

7 sin (x) - 4 = 0

What is the general solution for 2cos^2(x)+sin(x)-1=0,0<= x<= 2pi ?
The general solution for 2cos^2(x)+sin(x)-1=0,0<= x<= 2pi is x=(7pi)/6 ,x=(11pi)/6 ,x= pi/2