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

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

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

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

x = 2 πn

Degrees

x = 0^{\circ} + 36 0^{\circ} n

Solution steps

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

Subtract

1 + cos^{2} (x)

from both sides

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

= - 1 + 2 cos (x) - 1

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

Group like terms

2 cos (x) - 1 - 1 = 0

Subtract the numbers:

- 1 - 1 = - 2

2 cos (x) - 2 = 0

Move

2

to the right side

2 cos (x) - 2 = 0

Add

2

to both sides

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

Simplify

2 cos (x) = 2

2 cos (x) = 2

Divide both sides by

2

2 cos (x) = 2

Divide both sides by

2

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

Simplify

cos (x) = 1

cos (x) = 1

General solutions for

cos (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}

x = 0 + 2 πn

x = 0 + 2 πn

Solve

x = 0 + 2 πn : x = 2 πn

x = 0 + 2 πn

0 + 2 πn = 2 πn

x = 2 πn

x = 2 πn

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

4 tan^{2} (x) + 12 tan (x) - 27 = 0

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

cos^{4} (a) = 3 + 4 cos^{2} (a) + cos^{4} (a)

cos^{4} (t) = 1

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