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

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

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

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

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

Solution steps

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

Subtract

1

from both sides

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

Rewrite using trig identities

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

Solve by substitution

sin (x) = 1, sin (x) = - 1

sin (x) = 1 : x = \frac{π}{2} + 2 πn

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

Combine all the solutions

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

1 = sin (x) + 3 cos (x)

1 + sin (x) = 2 cos^{2} (x), 0 \leq x < 5

2 - 3 cos (x) = 5 + 3 cos (x)

sinh (x) = \frac{6}{5}

tan (2 x) tan (x) = 1

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