What is the general solution for 2cos^2(x)+sin(x)=5 ?

The general solution for 2cos^2(x)+sin(x)=5 is No Solution for x\in\mathbb{R}

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2cos^2(x)+sin(x)=5

Solution

2 cos^{2} (x) + sin (x) = 5

Solution

No Solution for x \in R

Solution steps

2 cos^{2} (x) + sin (x) = 5

Subtract

5

from both sides

2 cos^{2} (x) + sin (x) - 5 = 0

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

= - 5 + sin (x) + 2 (1 - sin^{2} (x))

Simplify

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

- 5 + sin (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)

= - 5 + sin (x) + 2 - 2 sin^{2} (x)

Simplify

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

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

Group like terms

= sin (x) - 2 sin^{2} (x) - 5 + 2

Add/Subtract the numbers:

- 5 + 2 = - 3

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

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

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

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

Solve by substitution

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

Let:

sin (x) = u

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

- 3 + u - 2 u^{2} = 0 : u = \frac{1}{4} - i \frac{23}{4}, u = \frac{1}{4} + i \frac{23}{4}

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

Write in the standard form

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

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

a = - 2, b = 1, c = - 3

u_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 ( - 2 ) ( - 3 )}{2 ( - 2 )}

u_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 ( - 2 ) ( - 3 )}{2 ( - 2 )}

Simplify

1^{2} - 4 (- 2) (- 3) : 23 i

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

Apply rule

1^{a} = 1

1^{2} = 1

= 1 - 4 (- 2) (- 3)

Apply rule

- (- a) = a

= 1 - 4 \cdot 2 \cdot 3

Multiply the numbers:

4 \cdot 2 \cdot 3 = 24

= 1 - 24

Subtract the numbers:

1 - 24 = - 23

= - 23

Apply radical rule:

- a = - 1 a

- 23 = - 1 23

= - 1 23

Apply imaginary number rule:

- 1 = i

= 23 i

u_{1, 2} = \frac{- 1 \pm 23 i}{2 ( - 2 )}

Separate the solutions

u_{1} = \frac{- 1 + 23 i}{2 ( - 2 )}, u_{2} = \frac{- 1 - 23 i}{2 ( - 2 )}

u = \frac{- 1 + 23 i}{2 ( - 2 )} : \frac{1}{4} - i \frac{23}{4}

\frac{- 1 + 23 i}{2 ( - 2 )}

Remove parentheses:

(- a) = - a

= \frac{- 1 + 23 i}{- 2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{- 1 + 23 i}{- 4}

Apply the fraction rule:

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

= - \frac{- 1 + 23 i}{4}

Rewrite

- \frac{- 1 + 23 i}{4}

in standard complex form:

\frac{1}{4} - \frac{23}{4} i

- \frac{- 1 + 23 i}{4}

Apply the fraction rule:

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

\frac{- 1 + 23 i}{4} = - (- \frac{1}{4}) - (\frac{23 i}{4})

= - (- \frac{1}{4}) - (\frac{23 i}{4})

Remove parentheses:

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

= \frac{1}{4} - \frac{23 i}{4}

= \frac{1}{4} - \frac{23}{4} i

u = \frac{- 1 - 23 i}{2 ( - 2 )} : \frac{1}{4} + i \frac{23}{4}

\frac{- 1 - 23 i}{2 ( - 2 )}

Remove parentheses:

(- a) = - a

= \frac{- 1 - 23 i}{- 2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{- 1 - 23 i}{- 4}

Apply the fraction rule:

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

= - \frac{- 1 - 23 i}{4}

Rewrite

- \frac{- 1 - 23 i}{4}

in standard complex form:

\frac{1}{4} + \frac{23}{4} i

- \frac{- 1 - 23 i}{4}

Apply the fraction rule:

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

\frac{- 1 - 23 i}{4} = - (- \frac{1}{4}) - (- \frac{23 i}{4})

= - (- \frac{1}{4}) - (- \frac{23 i}{4})

Apply rule

- (- a) = a

= \frac{1}{4} + \frac{23 i}{4}

= \frac{1}{4} + \frac{23}{4} i

The solutions to the quadratic equation are:

u = \frac{1}{4} - i \frac{23}{4}, u = \frac{1}{4} + i \frac{23}{4}

Substitute back

u = sin (x)

sin (x) = \frac{1}{4} - i \frac{23}{4}, sin (x) = \frac{1}{4} + i \frac{23}{4}

sin (x) = \frac{1}{4} - i \frac{23}{4}, sin (x) = \frac{1}{4} + i \frac{23}{4}

sin (x) = \frac{1}{4} - i \frac{23}{4} :

No Solution

sin (x) = \frac{1}{4} - i \frac{23}{4}

No Solution

sin (x) = \frac{1}{4} + i \frac{23}{4} :

No Solution

sin (x) = \frac{1}{4} + i \frac{23}{4}

No Solution

Combine all the solutions

No Solution for x \in R

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Frequently Asked Questions (FAQ)

What is the general solution for 2cos^2(x)+sin(x)=5 ?
The general solution for 2cos^2(x)+sin(x)=5 is No Solution for x\in\mathbb{R}