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

The general solution for 2+cos^2(x)=-5sin(x) is x=-0.57207…+2pin,x=pi+0.57207…+2pin

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Popular >

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

Solution

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

Solution

x = - 0.57207 \dots + 2 πn, x = π + 0.57207 \dots + 2 πn

Degrees

x = - 32.77771 \dots^{\circ} + 36 0^{\circ} n, x = 212.77771 \dots^{\circ} + 36 0^{\circ} n

Solution steps

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

Subtract

- 5 sin (x)

from both sides

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

Rewrite using trig identities

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

Simplify

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

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

Solve by substitution

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

Let:

sin (x) = u

3 - u^{2} + 5 u = 0

3 - u^{2} + 5 u = 0 : u = - \frac{- 5 + 37}{2}, u = \frac{5 + 37}{2}

3 - u^{2} + 5 u = 0

Write in the standard form

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

- u^{2} + 5 u + 3 = 0

Solve with the quadratic formula

- u^{2} + 5 u + 3 = 0

Quadratic Equation Formula:

For

a = - 1, b = 5, c = 3

u_{1, 2} = \frac{- 5 \pm 5 ^{2} - 4 ( - 1 ) \cdot 3}{2 ( - 1 )}

u_{1, 2} = \frac{- 5 \pm 5 ^{2} - 4 ( - 1 ) \cdot 3}{2 ( - 1 )}

5^{2} - 4 (- 1) \cdot 3 = 37

5^{2} - 4 (- 1) \cdot 3

Apply rule

- (- a) = a

= 5^{2} + 4 \cdot 1 \cdot 3

Multiply the numbers:

4 \cdot 1 \cdot 3 = 12

= 5^{2} + 12

5^{2} = 25

= 25 + 12

Add the numbers:

25 + 12 = 37

= 37

u_{1, 2} = \frac{- 5 \pm 37}{2 ( - 1 )}

Separate the solutions

u_{1} = \frac{- 5 + 37}{2 ( - 1 )}, u_{2} = \frac{- 5 - 37}{2 ( - 1 )}

u = \frac{- 5 + 37}{2 ( - 1 )} : - \frac{- 5 + 37}{2}

\frac{- 5 + 37}{2 ( - 1 )}

Remove parentheses:

(- a) = - a

= \frac{- 5 + 37}{- 2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{- 5 + 37}{- 2}

Apply the fraction rule:

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

= - \frac{- 5 + 37}{2}

u = \frac{- 5 - 37}{2 ( - 1 )} : \frac{5 + 37}{2}

\frac{- 5 - 37}{2 ( - 1 )}

Remove parentheses:

(- a) = - a

= \frac{- 5 - 37}{- 2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{- 5 - 37}{- 2}

Apply the fraction rule:

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

- 5 - 37 = - (5 + 37)

= \frac{5 + 37}{2}

The solutions to the quadratic equation are:

u = - \frac{- 5 + 37}{2}, u = \frac{5 + 37}{2}

Substitute back

u = sin (x)

sin (x) = - \frac{- 5 + 37}{2}, sin (x) = \frac{5 + 37}{2}

sin (x) = - \frac{- 5 + 37}{2}, sin (x) = \frac{5 + 37}{2}

sin (x) = - \frac{- 5 + 37}{2} : x = arcsin (- \frac{- 5 + 37}{2}) + 2 πn, x = π + arcsin (\frac{- 5 + 37}{2}) + 2 πn

sin (x) = - \frac{- 5 + 37}{2}

Apply trig inverse properties

sin (x) = - \frac{- 5 + 37}{2}

General solutions for

sin (x) = - \frac{- 5 + 37}{2}

sin (x) = - a \Rightarrow x = arcsin (- a) + 2 πn, x = π + arcsin (a) + 2 πn

x = arcsin (- \frac{- 5 + 37}{2}) + 2 πn, x = π + arcsin (\frac{- 5 + 37}{2}) + 2 πn

x = arcsin (- \frac{- 5 + 37}{2}) + 2 πn, x = π + arcsin (\frac{- 5 + 37}{2}) + 2 πn

sin (x) = \frac{5 + 37}{2} :

No Solution

sin (x) = \frac{5 + 37}{2}

- 1 \leq sin (x) \leq 1

No Solution

Combine all the solutions

x = arcsin (- \frac{- 5 + 37}{2}) + 2 πn, x = π + arcsin (\frac{- 5 + 37}{2}) + 2 πn

Show solutions in decimal form

x = - 0.57207 \dots + 2 πn, x = π + 0.57207 \dots + 2 πn

Popular Examples

tan^5(x)tan^2(x)=1

tan^{5} (x) tan^{2} (x) = 1

(sin^2(a)+1)/((1+tan^2(a)))=1

\frac{sin ^{2} ( a ) + 1}{( 1 + tan ^{2} ( a ) )} = 1

tan(a)= 5/9 ,b=6

tan (a) = \frac{5}{9}, b = 6

sin^3(x)sin^2(x)sin(x)=0

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

2cos^2(x)+1=cos^2(x)

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

Frequently Asked Questions (FAQ)

What is the general solution for 2+cos^2(x)=-5sin(x) ?
The general solution for 2+cos^2(x)=-5sin(x) is x=-0.57207…+2pin,x=pi+0.57207…+2pin