What is the general solution for 2sin^2(x)=5-5cos(x) ?

The general solution for 2sin^2(x)=5-5cos(x) is x=2pin

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2sin^2(x)=5-5cos(x)

Solution

2 sin^{2} (x) = 5 - 5 cos (x)

Solution

x = 2 πn

Degrees

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

Solution steps

2 sin^{2} (x) = 5 - 5 cos (x)

Subtract

5 - 5 cos (x)

from both sides

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

Rewrite using trig identities

- 5 + 2 sin^{2} (x) + 5 cos (x)

Use the Pythagorean identity:

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

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

= - 5 + 2 (1 - cos^{2} (x)) + 5 cos (x)

Simplify

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

- 5 + 2 (1 - cos^{2} (x)) + 5 cos (x)

Expand

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

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

Apply the distributive law:

a (b - c) = ab - a c

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

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

Multiply the numbers:

2 \cdot 1 = 2

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

= - 5 + 2 - 2 cos^{2} (x) + 5 cos (x)

Add/Subtract the numbers:

- 5 + 2 = - 3

= 5 cos (x) - 2 cos^{2} (x) - 3

= 5 cos (x) - 2 cos^{2} (x) - 3

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

Solve by substitution

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

Let:

cos (x) = u

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

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

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

Write in the standard form

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

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

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

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

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

5^{2} - 4 (- 2) (- 3) = 1

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

Apply rule

- (- a) = a

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

Multiply the numbers:

4 \cdot 2 \cdot 3 = 24

= 5^{2} - 24

5^{2} = 25

= 25 - 24

Subtract the numbers:

25 - 24 = 1

= 1

Apply rule

1 = 1

= 1

u_{1, 2} = \frac{- 5 \pm 1}{2 ( - 2 )}

Separate the solutions

u_{1} = \frac{- 5 + 1}{2 ( - 2 )}, u_{2} = \frac{- 5 - 1}{2 ( - 2 )}

u = \frac{- 5 + 1}{2 ( - 2 )} : 1

\frac{- 5 + 1}{2 ( - 2 )}

Remove parentheses:

(- a) = - a

= \frac{- 5 + 1}{- 2 \cdot 2}

Add/Subtract the numbers:

- 5 + 1 = - 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{- 5 - 1}{2 ( - 2 )} : \frac{3}{2}

\frac{- 5 - 1}{2 ( - 2 )}

Remove parentheses:

(- a) = - a

= \frac{- 5 - 1}{- 2 \cdot 2}

Subtract the numbers:

- 5 - 1 = - 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 = 1, u = \frac{3}{2}

Substitute back

u = cos (x)

cos (x) = 1, cos (x) = \frac{3}{2}

cos (x) = 1, cos (x) = \frac{3}{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) = \frac{3}{2} :

No Solution

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

- 1 \leq cos (x) \leq 1

No Solution

Combine all the solutions

x = 2 πn

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cos (θ) = \frac{65}{65}

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solvefor x,-4sin(x)=-cos^2(x)+1 solve for

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Frequently Asked Questions (FAQ)

What is the general solution for 2sin^2(x)=5-5cos(x) ?
The general solution for 2sin^2(x)=5-5cos(x) is x=2pin