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

The general solution for -sqrt(2)sin(2x)=sin^2(x)+cos^2(x) is x=(5pi)/8+pin,x=(7pi)/8+pin

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

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

x = \frac{5 π}{8} + πn, x = \frac{7 π}{8} + 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{2}{2}

General solutions for

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

2 x = \frac{5 π}{4} + 2 πn, 2 x = \frac{7 π}{4} + 2 πn

Solve

2 x = \frac{5 π}{4} + 2 πn : x = \frac{5 π}{8} + πn

Solve

2 x = \frac{7 π}{4} + 2 πn : x = \frac{7 π}{8} + πn

x = \frac{5 π}{8} + πn, x = \frac{7 π}{8} + πn

cos (x) = 0.998482

3 sin (2 x) - sin (x) = 0

(- 6 cos (x) - 5 sin (x))^{2} + 11 sin^{2} (x) = 51

cos (5 x) - sin (2 x) - cos (x) = 0

cosh (4 x) = 0

What is the general solution for -sqrt(2)sin(2x)=sin^2(x)+cos^2(x) ?
The general solution for -sqrt(2)sin(2x)=sin^2(x)+cos^2(x) is x=(5pi)/8+pin,x=(7pi)/8+pin