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

The general solution for 2sin(2x)=sin^2(x)+cos^2(x) is x= pi/(12)+pin,x=(5pi)/(12)+pin

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

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

x = \frac{π}{12} + πn, x = \frac{5 π}{12} + πn

Solution steps

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

Subtract

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

from both sides

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

Rewrite using trig identities

2 sin (2 x) - 1 = 0

Move

1

to the right side

2 sin (2 x) = 1

Divide both sides by

2

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

General solutions for

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

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

Solve

2 x = \frac{π}{6} + 2 πn : x = \frac{π}{12} + πn

Solve

2 x = \frac{5 π}{6} + 2 πn : x = \frac{5 π}{12} + πn

x = \frac{π}{12} + πn, x = \frac{5 π}{12} + πn

sin (θ) = \frac{7}{12}

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

x, cos (3 x) = 1

sec (x) + tan (x) = 2

\frac{1}{cos ^{2} ( x )} = \frac{4}{3}

What is the general solution for 2sin(2x)=sin^2(x)+cos^2(x) ?
The general solution for 2sin(2x)=sin^2(x)+cos^2(x) is x= pi/(12)+pin,x=(5pi)/(12)+pin