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

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

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

Solution

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

Solution

x = 2 πn

Degrees

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

Solution steps

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

Subtract

3 (1 - cos (x))

from both sides

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

Simplify

1 - cos^{2} (x) - (1 - cos (x)) \cdot 3 : 3 cos (x) - cos^{2} (x) - 2

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

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

Expand

- 3 (1 - cos (x)) : - 3 + 3 cos (x)

- 3 (1 - cos (x))

Apply the distributive law:

a (b - c) = ab - a c

a = - 3, b = 1, c = cos (x)

= - 3 \cdot 1 - (- 3) cos (x)

Apply minus-plus rules

- (- a) = a

= - 3 \cdot 1 + 3 cos (x)

Multiply the numbers:

3 \cdot 1 = 3

= - 3 + 3 cos (x)

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

Simplify

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

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

Group like terms

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

Add/Subtract the numbers:

1 - 3 = - 2

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

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

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

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

Solve by substitution

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

Let:

cos (x) = u

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

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

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

Write in the standard form

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

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

a = - 1, b = 3, c = - 2

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

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

3^{2} - 4 (- 1) (- 2) = 1

3^{2} - 4 (- 1) (- 2)

Apply rule

- (- a) = a

= 3^{2} - 4 \cdot 1 \cdot 2

Multiply the numbers:

4 \cdot 1 \cdot 2 = 8

= 3^{2} - 8

3^{2} = 9

= 9 - 8

Subtract the numbers:

9 - 8 = 1

= 1

Apply rule

1 = 1

= 1

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

Separate the solutions

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

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

\frac{- 3 + 1}{2 ( - 1 )}

Remove parentheses:

(- a) = - a

= \frac{- 3 + 1}{- 2 \cdot 1}

Add/Subtract the numbers:

- 3 + 1 = - 2

= \frac{- 2}{- 2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{- 2}{- 2}

Apply the fraction rule:

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

= \frac{2}{2}

Apply rule

\frac{a}{a} = 1

= 1

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

\frac{- 3 - 1}{2 ( - 1 )}

Remove parentheses:

(- a) = - a

= \frac{- 3 - 1}{- 2 \cdot 1}

Subtract the numbers:

- 3 - 1 = - 4

= \frac{- 4}{- 2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{- 4}{- 2}

Apply the fraction rule:

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

= \frac{4}{2}

Divide the numbers:

\frac{4}{2} = 2

= 2

The solutions to the quadratic equation are:

u = 1, u = 2

Substitute back

u = cos (x)

cos (x) = 1, cos (x) = 2

cos (x) = 1, cos (x) = 2

cos (x) = 1 : x = 2 πn

cos (x) = 1

General solutions for

cos (x) = 1

cos (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} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{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} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

x = 0 + 2 πn

x = 0 + 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

cos (x) = 2 :

No Solution

cos (x) = 2

- 1 \leq cos (x) \leq 1

No Solution

Combine all the solutions

x = 2 πn

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sin(2x)=-1/2 ,0<= x<= 2pi

sin (2 x) = - \frac{1}{2}, 0 \leq x \leq 2 π

Frequently Asked Questions (FAQ)

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