What is the general solution for 3cos^2(x)=1-5sin(x) ?

The general solution for 3cos^2(x)=1-5sin(x) is x=-0.33983…+2pin,x=pi+0.33983…+2pin

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3cos^2(x)=1-5sin(x)

Solution

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

Solution

x = - 0.33983 \dots + 2 πn, x = π + 0.33983 \dots + 2 πn

Degrees

x = - 19.47122 \dots^{\circ} + 36 0^{\circ} n, x = 199.47122 \dots^{\circ} + 36 0^{\circ} n

Solution steps

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

Subtract

1 - 5 sin (x)

from both sides

3 cos^{2} (x) - 1 + 5 sin (x) = 0

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

Simplify

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

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

Expand

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

3 (1 - sin^{2} (x))

Apply the distributive law:

a (b - c) = ab - a c

a = 3, b = 1, c = sin^{2} (x)

= 3 \cdot 1 - 3 sin^{2} (x)

Multiply the numbers:

3 \cdot 1 = 3

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

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

Add/Subtract the numbers:

- 1 + 3 = 2

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

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

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

Solve by substitution

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

Let:

sin (x) = u

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

2 - 3 u^{2} + 5 u = 0 : u = - \frac{1}{3}, u = 2

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

Write in the standard form

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

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

a = - 3, b = 5, c = 2

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

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

5^{2} - 4 (- 3) \cdot 2 = 7

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

Apply rule

- (- a) = a

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

Multiply the numbers:

4 \cdot 3 \cdot 2 = 24

= 5^{2} + 24

5^{2} = 25

= 25 + 24

Add the numbers:

25 + 24 = 49

= 49

Factor the number:

49 = 7^{2}

= 7^{2}

Apply radical rule:

7^{2} = 7

= 7

u_{1, 2} = \frac{- 5 \pm 7}{2 ( - 3 )}

Separate the solutions

u_{1} = \frac{- 5 + 7}{2 ( - 3 )}, u_{2} = \frac{- 5 - 7}{2 ( - 3 )}

u = \frac{- 5 + 7}{2 ( - 3 )} : - \frac{1}{3}

\frac{- 5 + 7}{2 ( - 3 )}

Remove parentheses:

(- a) = - a

= \frac{- 5 + 7}{- 2 \cdot 3}

Add/Subtract the numbers:

- 5 + 7 = 2

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

Multiply the numbers:

2 \cdot 3 = 6

= \frac{2}{- 6}

Apply the fraction rule:

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

= - \frac{2}{6}

Cancel the common factor:

2

= - \frac{1}{3}

u = \frac{- 5 - 7}{2 ( - 3 )} : 2

\frac{- 5 - 7}{2 ( - 3 )}

Remove parentheses:

(- a) = - a

= \frac{- 5 - 7}{- 2 \cdot 3}

Subtract the numbers:

- 5 - 7 = - 12

= \frac{- 12}{- 2 \cdot 3}

Multiply the numbers:

2 \cdot 3 = 6

= \frac{- 12}{- 6}

Apply the fraction rule:

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

= \frac{12}{6}

Divide the numbers:

\frac{12}{6} = 2

= 2

The solutions to the quadratic equation are:

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

Substitute back

u = sin (x)

sin (x) = - \frac{1}{3}, sin (x) = 2

sin (x) = - \frac{1}{3}, sin (x) = 2

sin (x) = - \frac{1}{3} : x = arcsin (- \frac{1}{3}) + 2 πn, x = π + arcsin (\frac{1}{3}) + 2 πn

sin (x) = - \frac{1}{3}

Apply trig inverse properties

sin (x) = - \frac{1}{3}

General solutions for

sin (x) = - \frac{1}{3}

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

x = arcsin (- \frac{1}{3}) + 2 πn, x = π + arcsin (\frac{1}{3}) + 2 πn

x = arcsin (- \frac{1}{3}) + 2 πn, x = π + arcsin (\frac{1}{3}) + 2 πn

sin (x) = 2 :

No Solution

sin (x) = 2

- 1 \leq sin (x) \leq 1

No Solution

Combine all the solutions

x = arcsin (- \frac{1}{3}) + 2 πn, x = π + arcsin (\frac{1}{3}) + 2 πn

Show solutions in decimal form

x = - 0.33983 \dots + 2 πn, x = π + 0.33983 \dots + 2 πn

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Frequently Asked Questions (FAQ)

What is the general solution for 3cos^2(x)=1-5sin(x) ?
The general solution for 3cos^2(x)=1-5sin(x) is x=-0.33983…+2pin,x=pi+0.33983…+2pin