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

The general solution for -2cos^2(x)+sin(x)+1=0 is x= pi/6+2pin,x=(5pi)/6+2pin,x=(3pi)/2+2pin

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

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

x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn, x = \frac{3 π}{2} + 2 πn

Solution steps

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

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

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

Combine all the solutions

x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn, x = \frac{3 π}{2} + 2 πn

tan (x) + sec (x) = 1, 0 \leq x \leq 2 π

8 sin (θ) cos (θ) = 3

tan (x) = \frac{7}{9}

cos (8 x) = - 1

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

What is the general solution for -2cos^2(x)+sin(x)+1=0 ?
The general solution for -2cos^2(x)+sin(x)+1=0 is x= pi/6+2pin,x=(5pi)/6+2pin,x=(3pi)/2+2pin