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

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

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Popular >

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

Solution

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

Solution

x = 2 πn, x = π + 2 πn, x = \frac{3 π}{2} + 2 πn

Degrees

x = 0^{\circ} + 36 0^{\circ} n, x = 18 0^{\circ} + 36 0^{\circ} n, x = 27 0^{\circ} + 36 0^{\circ} n

Solution steps

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

Subtract

- 3 sin (x)

from both sides

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

Rewrite using trig identities

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

Use the Double Angle identity:

sin (2 x) = 2 sin (x) cos (x)

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

cos (x) \cdot 2 sin (x) cos (x) = 2 cos^{2} (x) sin (x)

cos (x) \cdot 2 sin (x) cos (x)

Apply exponent rule:

a^{b} \cdot a^{c} = a^{b + c}

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

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

Add the numbers:

1 + 1 = 2

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

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

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

Factor

3 sin (x) + 3 sin^{2} (x) + 2 cos^{2} (x) sin (x) : sin (x) (3 + 3 sin (x) + 2 cos^{2} (x))

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

Apply exponent rule:

a^{b + c} = a^{b} a^{c}

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

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

Factor out common term

sin (x)

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

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

Solving each part separately

sin (x) = 0 or 3 + 3 sin (x) + 2 cos^{2} (x) = 0

sin (x) = 0 : x = 2 πn, x = π + 2 πn

sin (x) = 0

General solutions for

sin (x) = 0

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 = 0 + 2 πn, x = π + 2 πn

x = 0 + 2 πn, x = π + 2 πn

Solve

x = 0 + 2 πn : x = 2 πn

x = 0 + 2 πn

0 + 2 πn = 2 πn

x = 2 πn

x = 2 πn, x = π + 2 πn

3 + 3 sin (x) + 2 cos^{2} (x) = 0 : x = \frac{3 π}{2} + 2 πn

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

Simplify

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

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

Expand

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

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

Apply the distributive law:

a (b - c) = ab - a c

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

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

Multiply the numbers:

2 \cdot 1 = 2

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

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

Add the numbers:

3 + 2 = 5

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

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

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

Solve by substitution

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

Let:

sin (x) = u

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

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

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

Write in the standard form

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

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

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

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

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

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

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

Apply rule

- (- a) = a

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

Multiply the numbers:

4 \cdot 2 \cdot 5 = 40

= 3^{2} + 40

3^{2} = 9

= 9 + 40

Add the numbers:

9 + 40 = 49

= 49

Factor the number:

49 = 7^{2}

= 7^{2}

Apply radical rule:

7^{2} = 7

= 7

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

Separate the solutions

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

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

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

Remove parentheses:

(- a) = - a

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

Add/Subtract the numbers:

- 3 + 7 = 4

= \frac{4}{- 2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{4}{- 4}

Apply the fraction rule:

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

= - \frac{4}{4}

Apply rule

\frac{a}{a} = 1

= - 1

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

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

Remove parentheses:

(- a) = - a

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

Subtract the numbers:

- 3 - 7 = - 10

= \frac{- 10}{- 2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{- 10}{- 4}

Apply the fraction rule:

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

= \frac{10}{4}

Cancel the common factor:

2

= \frac{5}{2}

The solutions to the quadratic equation are:

u = - 1, u = \frac{5}{2}

Substitute back

u = sin (x)

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

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

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

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

No Solution

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

- 1 \leq sin (x) \leq 1

No Solution

Combine all the solutions

x = \frac{3 π}{2} + 2 πn

Combine all the solutions

x = 2 πn, x = π + 2 πn, x = \frac{3 π}{2} + 2 πn

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Frequently Asked Questions (FAQ)

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