What is the general solution for 3cos^2(x)=9sin(x)+9 ?

The general solution for 3cos^2(x)=9sin(x)+9 is x=(3pi)/2+2pin

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3cos^2(x)=9sin(x)+9

Solution

3 cos^{2} (x) = 9 sin (x) + 9

Solution

x = \frac{3 π}{2} + 2 πn

Degrees

x = 27 0^{\circ} + 36 0^{\circ} n

Solution steps

3 cos^{2} (x) = 9 sin (x) + 9

Subtract

9 sin (x) + 9

from both sides

3 cos^{2} (x) - 9 sin (x) - 9 = 0

Rewrite using trig identities

- 9 + 3 cos^{2} (x) - 9 sin (x)

Use the Pythagorean identity:

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

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

= - 9 + 3 (1 - sin^{2} (x)) - 9 sin (x)

Simplify

- 9 + 3 (1 - sin^{2} (x)) - 9 sin (x) : - 3 sin^{2} (x) - 9 sin (x) - 6

- 9 + 3 (1 - sin^{2} (x)) - 9 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)

= - 9 + 3 - 3 sin^{2} (x) - 9 sin (x)

Add/Subtract the numbers:

- 9 + 3 = - 6

= - 3 sin^{2} (x) - 9 sin (x) - 6

= - 3 sin^{2} (x) - 9 sin (x) - 6

- 6 - 3 sin^{2} (x) - 9 sin (x) = 0

Solve by substitution

- 6 - 3 sin^{2} (x) - 9 sin (x) = 0

Let:

sin (x) = u

- 6 - 3 u^{2} - 9 u = 0

- 6 - 3 u^{2} - 9 u = 0 : u = - 2, u = - 1

- 6 - 3 u^{2} - 9 u = 0

Write in the standard form

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

- 3 u^{2} - 9 u - 6 = 0

Solve with the quadratic formula

- 3 u^{2} - 9 u - 6 = 0

Quadratic Equation Formula:

For

a = - 3, b = - 9, c = - 6

u_{1, 2} = \frac{- ( - 9 ) \pm ( - 9 ) ^{2} - 4 ( - 3 ) ( - 6 )}{2 ( - 3 )}

u_{1, 2} = \frac{- ( - 9 ) \pm ( - 9 ) ^{2} - 4 ( - 3 ) ( - 6 )}{2 ( - 3 )}

(- 9)^{2} - 4 (- 3) (- 6) = 3

(- 9)^{2} - 4 (- 3) (- 6)

Apply rule

- (- a) = a

= (- 9)^{2} - 4 \cdot 3 \cdot 6

Apply exponent rule:

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

n

is even

(- 9)^{2} = 9^{2}

= 9^{2} - 4 \cdot 3 \cdot 6

Multiply the numbers:

4 \cdot 3 \cdot 6 = 72

= 9^{2} - 72

9^{2} = 81

= 81 - 72

Subtract the numbers:

81 - 72 = 9

= 9

Factor the number:

9 = 3^{2}

= 3^{2}

Apply radical rule:

3^{2} = 3

= 3

u_{1, 2} = \frac{- ( - 9 ) \pm 3}{2 ( - 3 )}

Separate the solutions

u_{1} = \frac{- ( - 9 ) + 3}{2 ( - 3 )}, u_{2} = \frac{- ( - 9 ) - 3}{2 ( - 3 )}

u = \frac{- ( - 9 ) + 3}{2 ( - 3 )} : - 2

\frac{- ( - 9 ) + 3}{2 ( - 3 )}

Remove parentheses:

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

= \frac{9 + 3}{- 2 \cdot 3}

Add the numbers:

9 + 3 = 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

u = \frac{- ( - 9 ) - 3}{2 ( - 3 )} : - 1

\frac{- ( - 9 ) - 3}{2 ( - 3 )}

Remove parentheses:

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

= \frac{9 - 3}{- 2 \cdot 3}

Subtract the numbers:

9 - 3 = 6

= \frac{6}{- 2 \cdot 3}

Multiply the numbers:

2 \cdot 3 = 6

= \frac{6}{- 6}

Apply the fraction rule:

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

= - \frac{6}{6}

Apply rule

\frac{a}{a} = 1

= - 1

The solutions to the quadratic equation are:

u = - 2, u = - 1

Substitute back

u = sin (x)

sin (x) = - 2, sin (x) = - 1

sin (x) = - 2, sin (x) = - 1

sin (x) = - 2 :

No Solution

sin (x) = - 2

- 1 \leq sin (x) \leq 1

No Solution

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

sin (x) = - 1

General solutions for

sin (x) = - 1

sin (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

x = \frac{3 π}{2} + 2 πn

x = \frac{3 π}{2} + 2 πn

Combine all the solutions

x = \frac{3 π}{2} + 2 πn

Popular Examples

cos^2(x)-2cos(x)=3

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

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

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

2sin^2(x)=1,+[0,+2pi]

2 sin^{2} (x) = 1, + [0, + 2 π]

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

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

sin(x)= 33/80

sin (x) = \frac{33}{80}

Frequently Asked Questions (FAQ)

What is the general solution for 3cos^2(x)=9sin(x)+9 ?
The general solution for 3cos^2(x)=9sin(x)+9 is x=(3pi)/2+2pin