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

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

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

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

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

Solution steps

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

Rewrite using trig identities

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

Solve by substitution

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

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

sin (x) = 0 : x = 2 πn, x = π + 2 πn

Combine all the solutions

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

8 sin (x) + 7 cos (x) = 1

sec (y) = 12

- sin (x) + 3 cos (x - \frac{π}{2}) = 1

cos (3 0^{\circ} + u) = 0.8177

cos (x - \frac{3 π}{2}) = \frac{1}{2}

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