What is the general solution for 3cos^2(x)+sin(x)+1=0 ?

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

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3cos^2(x)+sin(x)+1=0

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

x = \frac{3 π}{2} + 2 πn

Solution steps

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

Rewrite using trig identities

4 + sin (x) - 3 sin^{2} (x) = 0

Solve by substitution

sin (x) = - 1, sin (x) = \frac{4}{3}

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

sin (x) = \frac{4}{3} :

No Solution

Combine all the solutions

x = \frac{3 π}{2} + 2 πn

2 sin^{2} (\frac{1}{2} x) + C = 2 cos (\frac{x}{2}) sin (\frac{x}{2}) + C

sin (x) = 7

sin^{2} (x) cos^{2} (x) = \frac{2 - 2}{16}

(sin (x) - 1) (sin (x) - 5) = 0

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

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