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

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

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Popular >

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

Solution

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

Solution

No Solution for x \in R

Solution steps

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

Subtract

sin^{2} (x)

from both sides

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

Square both sides

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

Subtract

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

from both sides

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

Apply exponent rule:

a^{b} = a^{2} a^{b - 2}

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

Simplify

- 4 - sin^{4} (x) + 4 sin^{2} (x) + (1 - sin^{2} (x)) (1 - sin^{2} (x)) : 2 sin^{2} (x) - 3

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

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

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

Apply exponent rule:

a^{b} \cdot a^{c} = a^{b + c}

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

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

Add the numbers:

1 + 1 = 2

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

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

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

Apply Perfect Square Formula:

(a - b)^{2} = a^{2} - 2 ab + b^{2}

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

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

Simplify

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

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

Apply rule

1^{a} = 1

1^{2} = 1

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

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

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

Multiply the numbers:

2 \cdot 1 = 2

= 2 sin^{2} (x)

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

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

Apply exponent rule:

(a^{b})^{c} = a^{b c}

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

Multiply the numbers:

2 \cdot 2 = 4

= sin^{4} (x)

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

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

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

Simplify

- 4 - sin^{4} (x) + 4 sin^{2} (x) + 1 - 2 sin^{2} (x) + sin^{4} (x) : 2 sin^{2} (x) - 3

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

Group like terms

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

Add similar elements:

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

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

Add similar elements:

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

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

Add/Subtract the numbers:

- 4 + 1 = - 3

= 2 sin^{2} (x) - 3

= 2 sin^{2} (x) - 3

= 2 sin^{2} (x) - 3

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

Solve by substitution

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

Let:

sin (x) = u

- 3 + 2 u^{2} = 0

- 3 + 2 u^{2} = 0 : u = \frac{3}{2}, u = - \frac{3}{2}

- 3 + 2 u^{2} = 0

Move

3

to the right side

- 3 + 2 u^{2} = 0

Add

3

to both sides

- 3 + 2 u^{2} + 3 = 0 + 3

Simplify

2 u^{2} = 3

2 u^{2} = 3

Divide both sides by

2

2 u^{2} = 3

Divide both sides by

2

\frac{2 u ^{2}}{2} = \frac{3}{2}

Simplify

u^{2} = \frac{3}{2}

u^{2} = \frac{3}{2}

For

x^{2} = f (a)

the solutions are

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

u = \frac{3}{2}, u = - \frac{3}{2}

Substitute back

u = sin (x)

sin (x) = \frac{3}{2}, sin (x) = - \frac{3}{2}

sin (x) = \frac{3}{2}, sin (x) = - \frac{3}{2}

sin (x) = \frac{3}{2} :

No Solution

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

- 1 \leq sin (x) \leq 1

No Solution

sin (x) = - \frac{3}{2} :

No Solution

sin (x) = - \frac{3}{2}

- 1 \leq sin (x) \leq 1

No Solution

Combine all the solutions

No Solution

Verify solutions by plugging them into the original equation

Check the solutions by plugging them into

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

Remove the ones that don't agree with the equation.

No Solution for x \in R

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Frequently Asked Questions (FAQ)

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