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

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

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

Solution

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

Solution

x = 2 πn, x = π + 2 πn

Degrees

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

Solution steps

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

Subtract

1 - 3 sin (x)

from both sides

cos (2 x) - 1 + 3 sin (x) = 0

Rewrite using trig identities

- 1 + cos (2 x) + 3 sin (x)

Use the Double Angle identity:

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

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

Simplify

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

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

Solve by substitution

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

Let:

sin (x) = u

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

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

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

a = - 2, b = 3, c = 0

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

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

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

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

Apply rule

- (- a) = a

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

Separate the solutions

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

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

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

Remove parentheses:

(- a) = - a

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

Add/Subtract the numbers:

- 3 + 3 = 0

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

Multiply the numbers:

2 \cdot 2 = 4

= \frac{0}{- 4}

Apply the fraction rule:

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

= - \frac{0}{4}

Apply rule

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

= - 0

= 0

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

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

Remove parentheses:

(- a) = - a

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

Subtract the numbers:

- 3 - 3 = - 6

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

Multiply the numbers:

2 \cdot 2 = 4

= \frac{- 6}{- 4}

Apply the fraction rule:

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

= \frac{6}{4}

Cancel the common factor:

2

= \frac{3}{2}

The solutions to the quadratic equation are:

u = 0, u = \frac{3}{2}

Substitute back

u = sin (x)

sin (x) = 0, sin (x) = \frac{3}{2}

sin (x) = 0, sin (x) = \frac{3}{2}

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

sin (x) = \frac{3}{2} :

No Solution

sin (x) = \frac{3}{2}

- 1 \leq sin (x) \leq 1

No Solution

Combine all the solutions

x = 2 πn, x = π + 2 πn

Popular Examples

3sin^2(x)-7sin(x)-2=0

3 sin^{2} (x) - 7 sin (x) - 2 = 0

3sin(2t)+4=1,-pi/2 <= t<= pi/2

3 sin (2 t) + 4 = 1, - \frac{π}{2} \leq t \leq \frac{π}{2}

sin^2(z)=-1

sin^{2} (z) = - 1

2cos(x)+sin(x)=1

2 cos (x) + sin (x) = 1

(sin(x))/(12)=(sin(42))/(22)

\frac{sin ( x )}{12} = \frac{sin ( 4 2 ^{\circ} )}{22}

Frequently Asked Questions (FAQ)

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