What is the general solution for 5-5cos(x)=3sin^2(x) ?

The general solution for 5-5cos(x)=3sin^2(x) is x=2pin,x=0.84106…+2pin,x=2pi-0.84106…+2pin

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5-5cos(x)=3sin^2(x)

Solution

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

Solution

x = 2 πn, x = 0.84106 \dots + 2 πn, x = 2 π - 0.84106 \dots + 2 πn

Degrees

x = 0^{\circ} + 36 0^{\circ} n, x = 48.18968 \dots^{\circ} + 36 0^{\circ} n, x = 311.81031 \dots^{\circ} + 36 0^{\circ} n

Solution steps

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

Subtract

3 sin^{2} (x)

from both sides

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

Rewrite using trig identities

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

Simplify

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

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

Expand

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

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

Apply the distributive law:

a (b - c) = ab - a c

a = - 3, b = 1, c = cos^{2} (x)

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

Apply minus-plus rules

- (- a) = a

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

Multiply the numbers:

3 \cdot 1 = 3

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

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

Subtract the numbers:

5 - 3 = 2

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

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

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

Solve by substitution

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

Let:

cos (x) = u

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

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

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

Write in the standard form

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

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

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

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

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

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

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

Apply exponent rule:

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

n

is even

(- 5)^{2} = 5^{2}

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

Multiply the numbers:

4 \cdot 3 \cdot 2 = 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 \cdot 3}

Separate the solutions

u_{1} = \frac{- ( - 5 ) + 1}{2 \cdot 3}, u_{2} = \frac{- ( - 5 ) - 1}{2 \cdot 3}

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

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

Apply rule

- (- a) = a

= \frac{5 + 1}{2 \cdot 3}

Add the numbers:

5 + 1 = 6

= \frac{6}{2 \cdot 3}

Multiply the numbers:

2 \cdot 3 = 6

= \frac{6}{6}

Apply rule

\frac{a}{a} = 1

= 1

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

\frac{- ( - 5 ) - 1}{2 \cdot 3}

Apply rule

- (- a) = a

= \frac{5 - 1}{2 \cdot 3}

Subtract the numbers:

5 - 1 = 4

= \frac{4}{2 \cdot 3}

Multiply the numbers:

2 \cdot 3 = 6

= \frac{4}{6}

Cancel the common factor:

2

= \frac{2}{3}

The solutions to the quadratic equation are:

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

Substitute back

u = cos (x)

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

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

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{2}{3} : x = arccos (\frac{2}{3}) + 2 πn, x = 2 π - arccos (\frac{2}{3}) + 2 πn

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

Apply trig inverse properties

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

General solutions for

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

cos (x) = a \Rightarrow x = arccos (a) + 2 πn, x = 2 π - arccos (a) + 2 πn

x = arccos (\frac{2}{3}) + 2 πn, x = 2 π - arccos (\frac{2}{3}) + 2 πn

x = arccos (\frac{2}{3}) + 2 πn, x = 2 π - arccos (\frac{2}{3}) + 2 πn

Combine all the solutions

x = 2 πn, x = arccos (\frac{2}{3}) + 2 πn, x = 2 π - arccos (\frac{2}{3}) + 2 πn

Show solutions in decimal form

x = 2 πn, x = 0.84106 \dots + 2 πn, x = 2 π - 0.84106 \dots + 2 πn

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Frequently Asked Questions (FAQ)

What is the general solution for 5-5cos(x)=3sin^2(x) ?
The general solution for 5-5cos(x)=3sin^2(x) is x=2pin,x=0.84106…+2pin,x=2pi-0.84106…+2pin