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

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

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

Solution

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

Solution

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

Degrees

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

Solution steps

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

Subtract

3 sin (x)

from both sides

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

Rewrite using trig identities

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

Use the Double Angle identity:

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

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

Simplify

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

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

Add similar elements:

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

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

- 1 + 1 = 0

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

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

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

Solve by substitution

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

Let:

sin (x) = u

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

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

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

Write in the standard form

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

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

a = - 3, b = - 3, c = 0

u_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 ( - 3 ) \cdot 0}{2 ( - 3 )}

u_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 ( - 3 ) \cdot 0}{2 ( - 3 )}

(- 3)^{2} - 4 (- 3) \cdot 0 = 3

(- 3)^{2} - 4 (- 3) \cdot 0

Apply rule

- (- a) = a

= (- 3)^{2} + 4 \cdot 3 \cdot 0

Apply exponent rule:

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

n

is even

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

= 3^{2} + 4 \cdot 3 \cdot 0

Apply rule

0 \cdot a = 0

= 3^{2} + 0

3^{2} + 0 = 3^{2}

= 3^{2}

Apply radical rule:

assuming

a \geq 0

= 3

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

Separate the solutions

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

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

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

Remove parentheses:

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

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

Add the numbers:

3 + 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

u = \frac{- ( - 3 ) - 3}{2 ( - 3 )} : 0

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

Remove parentheses:

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

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

Subtract the numbers:

3 - 3 = 0

= \frac{0}{- 2 \cdot 3}

Multiply the numbers:

2 \cdot 3 = 6

= \frac{0}{- 6}

Apply the fraction rule:

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

= - \frac{0}{6}

Apply rule

\frac{0}{a} = 0, a \neq = 0

= - 0

= 0

The solutions to the quadratic equation are:

u = - 1, u = 0

Substitute back

u = sin (x)

sin (x) = - 1, sin (x) = 0

sin (x) = - 1, sin (x) = 0

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) = 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

Combine all the solutions

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

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Frequently Asked Questions (FAQ)

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