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

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

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

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

x = 2 πn, x = π + 2 πn

Degrees

x = 0^{\circ} + 36 0^{\circ} n, x = 18 0^{\circ} + 36 0^{\circ} n

Solution steps

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

Subtract

1 - sin^{2} (x)

from both sides

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

Simplify

- 2 + sin^{2} (x) + 2 (1 - sin^{2} (x)) : - sin^{2} (x)

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

Expand

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

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

Apply the distributive law:

a (b - c) = ab - a c

a = 2, b = 1, c = sin^{2} (x)

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

Multiply the numbers:

2 \cdot 1 = 2

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

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

Simplify

- 2 + sin^{2} (x) + 2 - 2 sin^{2} (x) : - sin^{2} (x)

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

Group like terms

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

Add similar elements:

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

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

- 2 + 2 = 0

= - sin^{2} (x)

= - sin^{2} (x)

= - sin^{2} (x)

- sin^{2} (x) = 0

Divide both sides by

- 1

- sin^{2} (x) = 0

Divide both sides by

- 1

- sin^{2} (x) = 0

Divide both sides by

- 1

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

Simplify

sin^{2} (x) = 0

sin^{2} (x) = 0

Apply rule

x^{n} = 0 \Rightarrow x = 0

sin (x) = 0

General solutions for

sin (x) = 0

sin (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} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{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} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

x = 0 + 2 πn, x = π + 2 πn

x = 0 + 2 πn, x = π + 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, x = π + 2 πn

cos (x) + 0.65 = 0

sin^{2} (x) = 0.12

6 cos (x) = - 32

cos (x) = \frac{7}{12}

5 tan (x) + 3 = 4 (1 + tan (x))

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