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=2pin,x=pi+2pin,x= pi/6+2pin,x=(5pi)/6+2pin

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

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

x = 2 πn, x = π + 2 πn, x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 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) = 0, sin (x) = \frac{1}{2}

sin (x) = 0 : x = 2 πn, x = π + 2 πn

sin (x) = \frac{1}{2} : x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

Combine all the solutions

x = 2 πn, x = π + 2 πn, x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

cos (3 π)

8 sin^{2} (x) + 2 cos (x) = 7

tan (\frac{19 π}{12})

\frac{1}{sec ( x ) - tan ( x )} = sec (x) + tan (x)

cos (9 0^{\circ})

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=2pin,x=pi+2pin,x= pi/6+2pin,x=(5pi)/6+2pin