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

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

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

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

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

Solution steps

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

Subtract

sin^{2} (x) + 2 sin (x)

from both sides

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

Rewrite using trig identities

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

Solve by substitution

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

sin (x) = - 1 : x = \frac{3 π}{2} + 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 = \frac{3 π}{2} + 2 πn, x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

arctan (x) = 26

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

\frac{sin ( 3 1 ^{\circ} )}{12} = \frac{sin ( x )}{16}

tan (θ) = - \frac{8}{6}

cos^{2} (θ) - \frac{3}{2} = \frac{5}{2} cos (θ)

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