What is the general solution for 4cos^2(x)=4-2sin^2(x) ?

The general solution for 4cos^2(x)=4-2sin^2(x) is x=2pin,x=pi+2pin

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4cos^2(x)=4-2sin^2(x)

4 cos^{2} (x) = 4 - 2 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

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

Subtract

4 - 2 sin^{2} (x)

from both sides

4 cos^{2} (x) - 4 + 2 sin^{2} (x) = 0

Rewrite using trig identities

- 4 + 2 sin^{2} (x) + 4 cos^{2} (x)

Use the Pythagorean identity:

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

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

= - 4 + 2 sin^{2} (x) + 4 (1 - sin^{2} (x))

Simplify

- 4 + 2 sin^{2} (x) + 4 (1 - sin^{2} (x)) : - 2 sin^{2} (x)

- 4 + 2 sin^{2} (x) + 4 (1 - sin^{2} (x))

Expand

4 (1 - sin^{2} (x)) : 4 - 4 sin^{2} (x)

4 (1 - sin^{2} (x))

Apply the distributive law:

a (b - c) = ab - a c

a = 4, b = 1, c = sin^{2} (x)

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

Multiply the numbers:

4 \cdot 1 = 4

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

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

Simplify

- 4 + 2 sin^{2} (x) + 4 - 4 sin^{2} (x) : - 2 sin^{2} (x)

- 4 + 2 sin^{2} (x) + 4 - 4 sin^{2} (x)

Group like terms

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

Add similar elements:

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

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

- 4 + 4 = 0

= - 2 sin^{2} (x)

= - 2 sin^{2} (x)

= - 2 sin^{2} (x)

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

Divide both sides by

- 2

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

Divide both sides by

- 2

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

Divide both sides by

- 2

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

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

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

2 sin (x) + \frac{1}{sin ( x )} = 3

cos (x - \frac{π}{4}) = \frac{3}{2}

sec (θ) = 3, sin (θ) < 0

cos (x) tan (x) - cos (x) = 0, 0 \leq x \leq 2 π

What is the general solution for 4cos^2(x)=4-2sin^2(x) ?
The general solution for 4cos^2(x)=4-2sin^2(x) is x=2pin,x=pi+2pin