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

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

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

Solution

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

Solution

No Solution for x \in R

Solution steps

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

Subtract

2 - cos (x)

from both sides

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

Simplify

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

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

Group like terms

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

Add/Subtract the numbers:

- 2 + 1 = - 1

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

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

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

Solve by substitution

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

Let:

cos (x) = u

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

- 1 + u - u^{2} = 0 : u = \frac{1}{2} - i \frac{3}{2}, u = \frac{1}{2} + i \frac{3}{2}

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

Write in the standard form

a x^{2} + b x + c = 0

- u^{2} + u - 1 = 0

Solve with the quadratic formula

- u^{2} + u - 1 = 0

Quadratic Equation Formula:

For

a = - 1, b = 1, c = - 1

u_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 ( - 1 ) ( - 1 )}{2 ( - 1 )}

u_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 ( - 1 ) ( - 1 )}{2 ( - 1 )}

Simplify

1^{2} - 4 (- 1) (- 1) : 3 i

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

Apply rule

1^{a} = 1

1^{2} = 1

= 1 - 4 (- 1) (- 1)

Apply rule

- (- a) = a

= 1 - 4 \cdot 1 \cdot 1

Multiply the numbers:

4 \cdot 1 \cdot 1 = 4

= 1 - 4

Subtract the numbers:

1 - 4 = - 3

= - 3

Apply radical rule:

- a = - 1 a

- 3 = - 1 3

= - 1 3

Apply imaginary number rule:

- 1 = i

= 3 i

u_{1, 2} = \frac{- 1 \pm 3 i}{2 ( - 1 )}

Separate the solutions

u_{1} = \frac{- 1 + 3 i}{2 ( - 1 )}, u_{2} = \frac{- 1 - 3 i}{2 ( - 1 )}

u = \frac{- 1 + 3 i}{2 ( - 1 )} : \frac{1}{2} - i \frac{3}{2}

\frac{- 1 + 3 i}{2 ( - 1 )}

Remove parentheses:

(- a) = - a

= \frac{- 1 + 3 i}{- 2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{- 1 + 3 i}{- 2}

Apply the fraction rule:

\frac{a}{- b} = - \frac{a}{b}

= - \frac{- 1 + 3 i}{2}

Rewrite

- \frac{- 1 + 3 i}{2}

in standard complex form:

\frac{1}{2} - \frac{3}{2} i

- \frac{- 1 + 3 i}{2}

Apply the fraction rule:

\frac{a \pm b}{c} = \frac{a}{c} \pm \frac{b}{c}

\frac{- 1 + 3 i}{2} = - (- \frac{1}{2}) - (\frac{3 i}{2})

= - (- \frac{1}{2}) - (\frac{3 i}{2})

Remove parentheses:

(a) = a, - (- a) = a

= \frac{1}{2} - \frac{3 i}{2}

= \frac{1}{2} - \frac{3}{2} i

u = \frac{- 1 - 3 i}{2 ( - 1 )} : \frac{1}{2} + i \frac{3}{2}

\frac{- 1 - 3 i}{2 ( - 1 )}

Remove parentheses:

(- a) = - a

= \frac{- 1 - 3 i}{- 2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{- 1 - 3 i}{- 2}

Apply the fraction rule:

\frac{a}{- b} = - \frac{a}{b}

= - \frac{- 1 - 3 i}{2}

Rewrite

- \frac{- 1 - 3 i}{2}

in standard complex form:

\frac{1}{2} + \frac{3}{2} i

- \frac{- 1 - 3 i}{2}

Apply the fraction rule:

\frac{a \pm b}{c} = \frac{a}{c} \pm \frac{b}{c}

\frac{- 1 - 3 i}{2} = - (- \frac{1}{2}) - (- \frac{3 i}{2})

= - (- \frac{1}{2}) - (- \frac{3 i}{2})

Apply rule

- (- a) = a

= \frac{1}{2} + \frac{3 i}{2}

= \frac{1}{2} + \frac{3}{2} i

The solutions to the quadratic equation are:

u = \frac{1}{2} - i \frac{3}{2}, u = \frac{1}{2} + i \frac{3}{2}

Substitute back

u = cos (x)

cos (x) = \frac{1}{2} - i \frac{3}{2}, cos (x) = \frac{1}{2} + i \frac{3}{2}

cos (x) = \frac{1}{2} - i \frac{3}{2}, cos (x) = \frac{1}{2} + i \frac{3}{2}

cos (x) = \frac{1}{2} - i \frac{3}{2} :

No Solution

cos (x) = \frac{1}{2} - i \frac{3}{2}

No Solution

cos (x) = \frac{1}{2} + i \frac{3}{2} :

No Solution

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

No Solution

Combine all the solutions

No Solution for x \in R

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2cos^2(x)-cos(x)=2-sec(x)

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Frequently Asked Questions (FAQ)

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