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

The general solution for sin^2(x)=4-2cos^2(x) is No Solution for x\in\mathbb{R}

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Popular >

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

Solution

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

Solution

No Solution for x \in R

Solution steps

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

Subtract

4 - 2 cos^{2} (x)

from both sides

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

Rewrite using trig identities

- 4 + 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)

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

Simplify

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

- 4 + 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)

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

Simplify

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

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

Group like terms

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

Add similar elements:

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

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

Add/Subtract the numbers:

- 4 + 2 = - 2

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

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

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

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

Solve by substitution

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

Let:

sin (x) = u

- 2 - u^{2} = 0

- 2 - u^{2} = 0 : u = 2 i, u = - 2 i

- 2 - u^{2} = 0

Move

2

to the right side

- 2 - u^{2} = 0

Add

2

to both sides

- 2 - u^{2} + 2 = 0 + 2

Simplify

- u^{2} = 2

- u^{2} = 2

Divide both sides by

- 1

- u^{2} = 2

Divide both sides by

- 1

\frac{- u ^{2}}{- 1} = \frac{2}{- 1}

Simplify

u^{2} = - 2

u^{2} = - 2

For

x^{2} = f (a)

the solutions are

x = f (a), - f (a)

u = - 2, u = - - 2

Simplify

- 2 : 2 i

- 2

Apply radical rule:

- a = - 1 a

- 2 = - 1 2

= - 1 2

Apply imaginary number rule:

- 1 = i

= 2 i

Simplify

- - 2 : - 2 i

- - 2

Simplify

- 2 : 2 i

- 2

Apply radical rule:

- a = - 1 a

- 2 = - 1 2

= - 1 2

Apply imaginary number rule:

- 1 = i

= 2 i

= - 2 i

u = 2 i, u = - 2 i

Substitute back

u = sin (x)

sin (x) = 2 i, sin (x) = - 2 i

sin (x) = 2 i, sin (x) = - 2 i

sin (x) = 2 i :

No Solution

sin (x) = 2 i

No Solution

sin (x) = - 2 i :

No Solution

sin (x) = - 2 i

No Solution

Combine all the solutions

No Solution for x \in R

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12sin^2(x)-13sin(x)+3=0

12 sin^{2} (x) - 13 sin (x) + 3 = 0

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sin (θ) = \frac{1}{2}

csc^2(x)-1=0

csc^{2} (x) - 1 = 0

Frequently Asked Questions (FAQ)

What is the general solution for sin^2(x)=4-2cos^2(x) ?
The general solution for sin^2(x)=4-2cos^2(x) is No Solution for x\in\mathbb{R}