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

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

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cos^2(x/2)-sin^2(x/2)=1-sin(x)

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

x = 2 πn, x = 2 πn + \frac{π}{2}

Solution steps

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

Subtract

1 - sin (x)

from both sides

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

Rewrite using trig identities

- 1 + 2 sin (x + \frac{π}{4}) = 0

Move

1

to the right side

2 sin (x + \frac{π}{4}) = 1

Divide both sides by

2

sin (x + \frac{π}{4}) = \frac{2}{2}

General solutions for

sin (x + \frac{π}{4}) = \frac{2}{2}

x + \frac{π}{4} = \frac{π}{4} + 2 πn, x + \frac{π}{4} = \frac{3 π}{4} + 2 πn

Solve

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

Solve

x + \frac{π}{4} = \frac{3 π}{4} + 2 πn : x = 2 πn + \frac{π}{2}

x = 2 πn, x = 2 πn + \frac{π}{2}

- 2 + 3 csc (x) = 5

2 sin (2 x) = sin (4 x), 0 \leq x \leq π

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

- a c_{1} sin (a x) + a c_{2} cos (a x) = 0

3 - 3 cos (x) = 0

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