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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

Solution

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

Solution

No Solution for x \in R

Solution steps

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

Subtract

sin (x)

from both sides

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

Simplify

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

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

Solve by substitution

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

Let:

sin (x) = u

3 - u - u^{2} = 0

3 - u - u^{2} = 0 : u = - \frac{1 + 13}{2}, u = \frac{13 - 1}{2}

3 - u - u^{2} = 0

Write in the standard form

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

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

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

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

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

(- 1)^{2} - 4 (- 1) \cdot 3 = 13

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

Apply rule

- (- a) = a

= (- 1)^{2} + 4 \cdot 1 \cdot 3

(- 1)^{2} = 1

(- 1)^{2}

Apply exponent rule:

(- a)^{n} = a^{n},

n

is even

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

= 1^{2}

Apply rule

1^{a} = 1

= 1

4 \cdot 1 \cdot 3 = 12

4 \cdot 1 \cdot 3

Multiply the numbers:

4 \cdot 1 \cdot 3 = 12

= 12

= 1 + 12

Add the numbers:

1 + 12 = 13

= 13

u_{1, 2} = \frac{- ( - 1 ) \pm 13}{2 ( - 1 )}

Separate the solutions

u_{1} = \frac{- ( - 1 ) + 13}{2 ( - 1 )}, u_{2} = \frac{- ( - 1 ) - 13}{2 ( - 1 )}

u = \frac{- ( - 1 ) + 13}{2 ( - 1 )} : - \frac{1 + 13}{2}

\frac{- ( - 1 ) + 13}{2 ( - 1 )}

Remove parentheses:

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

= \frac{1 + 13}{- 2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{1 + 13}{- 2}

Apply the fraction rule:

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

= - \frac{1 + 13}{2}

u = \frac{- ( - 1 ) - 13}{2 ( - 1 )} : \frac{13 - 1}{2}

\frac{- ( - 1 ) - 13}{2 ( - 1 )}

Remove parentheses:

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

= \frac{1 - 13}{- 2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{1 - 13}{- 2}

Apply the fraction rule:

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

1 - 13 = - (13 - 1)

= \frac{13 - 1}{2}

The solutions to the quadratic equation are:

u = - \frac{1 + 13}{2}, u = \frac{13 - 1}{2}

Substitute back

u = sin (x)

sin (x) = - \frac{1 + 13}{2}, sin (x) = \frac{13 - 1}{2}

sin (x) = - \frac{1 + 13}{2}, sin (x) = \frac{13 - 1}{2}

sin (x) = - \frac{1 + 13}{2} :

No Solution

sin (x) = - \frac{1 + 13}{2}

- 1 \leq sin (x) \leq 1

No Solution

sin (x) = \frac{13 - 1}{2} :

No Solution

sin (x) = \frac{13 - 1}{2}

- 1 \leq sin (x) \leq 1

No Solution

Combine all the solutions

No Solution for x \in R

Popular Examples

-sin(2x)-3cos(x)=0

- sin (2 x) - 3 cos (x) = 0

solvefor x,y=3cos(fxx+pi/2)+5 solve for

x, y = 3 cos (f xx + \frac{π}{2}) + 5

sin(x)cos(x)=sin(x),0<x<= 2pi

sin (x) cos (x) = sin (x), 0 < x \leq 2 π

cos^5(x)-sin(x)=0

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

solvefor x,y= 1/(5sin(2x))solve for

x, y = \frac{1}{5 sin ( 2 x )}

Frequently Asked Questions (FAQ)

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