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

The general solution for cos(2x)-sin^2(x)=cos^2(x) is x=pin

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

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

x = πn

Degrees

x = 0^{\circ} + 18 0^{\circ} n

Solution steps

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

Subtract

cos^{2} (x)

from both sides

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

= cos (2 x) - 1

cos (2 x) - 1 = 0

Move

1

to the right side

cos (2 x) - 1 = 0

Add

1

to both sides

cos (2 x) - 1 + 1 = 0 + 1

Simplify

cos (2 x) = 1

cos (2 x) = 1

General solutions for

cos (2 x) = 1

cos (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

2 x = 0 + 2 πn

2 x = 0 + 2 πn

Solve

2 x = 0 + 2 πn : x = πn

2 x = 0 + 2 πn

0 + 2 πn = 2 πn

2 x = 2 πn

Divide both sides by

2

2 x = 2 πn

Divide both sides by

2

\frac{2 x}{2} = \frac{2 πn}{2}

Simplify

x = πn

x = πn

x = πn

cos (x) = 0, 56

cos (\frac{x}{2}) = \frac{5}{6}

sin (B) = \frac{9}{16}

4 5^{\circ} = 57. 7^{\circ} + arctan (\frac{3.5}{x}) - arctan (\frac{175}{x})

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

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