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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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 = 147.22228 \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

= - sin^{2} (x) - 5 sin (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{37 - 5}{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

Apply exponent rule:

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

n

is even

(- 5)^{2} = 5^{2}

= 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, - (- 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{37 - 5}{2}

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

Remove parentheses:

(- a) = - a, - (- 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 = - (37 - 5)

= \frac{37 - 5}{2}

The solutions to the quadratic equation are:

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

Substitute back

u = sin (x)

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

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

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

No Solution

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

- 1 \leq sin (x) \leq 1

No Solution

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

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

Apply trig inverse properties

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

General solutions for

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

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

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

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

Combine all the solutions

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

Show solutions in decimal form

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

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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