What is the general solution for 3cos^2(x)-5cos(x)=1+3sin^2(x) ?

The general solution for 3cos^2(x)-5cos(x)=1+3sin^2(x) is x=(2pi)/3+2pin,x=(4pi)/3+2pin

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3cos^2(x)-5cos(x)=1+3sin^2(x)

Solution

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

Solution

x = \frac{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn

Degrees

x = 12 0^{\circ} + 36 0^{\circ} n, x = 24 0^{\circ} + 36 0^{\circ} n

Solution steps

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

Subtract

1 + 3 sin^{2} (x)

from both sides

3 cos^{2} (x) - 5 cos (x) - 1 - 3 sin^{2} (x) = 0

Rewrite using trig identities

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

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

Simplify

- 1 + 3 cos^{2} (x) - 3 (1 - cos^{2} (x)) - 5 cos (x) : 6 cos^{2} (x) - 5 cos (x) - 4

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

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

Simplify

- 1 + 3 cos^{2} (x) - 3 + 3 cos^{2} (x) - 5 cos (x) : 6 cos^{2} (x) - 5 cos (x) - 4

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

Group like terms

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

Add similar elements:

3 cos^{2} (x) + 3 cos^{2} (x) = 6 cos^{2} (x)

= 6 cos^{2} (x) - 5 cos (x) - 1 - 3

Subtract the numbers:

- 1 - 3 = - 4

= 6 cos^{2} (x) - 5 cos (x) - 4

= 6 cos^{2} (x) - 5 cos (x) - 4

= 6 cos^{2} (x) - 5 cos (x) - 4

- 4 - 5 cos (x) + 6 cos^{2} (x) = 0

Solve by substitution

- 4 - 5 cos (x) + 6 cos^{2} (x) = 0

Let:

cos (x) = u

- 4 - 5 u + 6 u^{2} = 0

- 4 - 5 u + 6 u^{2} = 0 : u = \frac{4}{3}, u = - \frac{1}{2}

- 4 - 5 u + 6 u^{2} = 0

Write in the standard form

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

6 u^{2} - 5 u - 4 = 0

Solve with the quadratic formula

6 u^{2} - 5 u - 4 = 0

Quadratic Equation Formula:

For

a = 6, b = - 5, c = - 4

u_{1, 2} = \frac{- ( - 5 ) \pm ( - 5 ) ^{2} - 4 \cdot 6 ( - 4 )}{2 \cdot 6}

u_{1, 2} = \frac{- ( - 5 ) \pm ( - 5 ) ^{2} - 4 \cdot 6 ( - 4 )}{2 \cdot 6}

(- 5)^{2} - 4 \cdot 6 (- 4) = 11

(- 5)^{2} - 4 \cdot 6 (- 4)

Apply rule

- (- a) = a

= (- 5)^{2} + 4 \cdot 6 \cdot 4

Apply exponent rule:

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

n

is even

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

= 5^{2} + 4 \cdot 6 \cdot 4

Multiply the numbers:

4 \cdot 6 \cdot 4 = 96

= 5^{2} + 96

5^{2} = 25

= 25 + 96

Add the numbers:

25 + 96 = 121

= 121

Factor the number:

121 = 1 1^{2}

= 1 1^{2}

Apply radical rule:

1 1^{2} = 11

= 11

u_{1, 2} = \frac{- ( - 5 ) \pm 11}{2 \cdot 6}

Separate the solutions

u_{1} = \frac{- ( - 5 ) + 11}{2 \cdot 6}, u_{2} = \frac{- ( - 5 ) - 11}{2 \cdot 6}

u = \frac{- ( - 5 ) + 11}{2 \cdot 6} : \frac{4}{3}

\frac{- ( - 5 ) + 11}{2 \cdot 6}

Apply rule

- (- a) = a

= \frac{5 + 11}{2 \cdot 6}

Add the numbers:

5 + 11 = 16

= \frac{16}{2 \cdot 6}

Multiply the numbers:

2 \cdot 6 = 12

= \frac{16}{12}

Cancel the common factor:

4

= \frac{4}{3}

u = \frac{- ( - 5 ) - 11}{2 \cdot 6} : - \frac{1}{2}

\frac{- ( - 5 ) - 11}{2 \cdot 6}

Apply rule

- (- a) = a

= \frac{5 - 11}{2 \cdot 6}

Subtract the numbers:

5 - 11 = - 6

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

Multiply the numbers:

2 \cdot 6 = 12

= \frac{- 6}{12}

Apply the fraction rule:

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

= - \frac{6}{12}

Cancel the common factor:

6

= - \frac{1}{2}

The solutions to the quadratic equation are:

u = \frac{4}{3}, u = - \frac{1}{2}

Substitute back

u = cos (x)

cos (x) = \frac{4}{3}, cos (x) = - \frac{1}{2}

cos (x) = \frac{4}{3}, cos (x) = - \frac{1}{2}

cos (x) = \frac{4}{3} :

No Solution

cos (x) = \frac{4}{3}

- 1 \leq cos (x) \leq 1

No Solution

cos (x) = - \frac{1}{2} : x = \frac{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn

cos (x) = - \frac{1}{2}

General solutions for

cos (x) = - \frac{1}{2}

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 = \frac{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn

x = \frac{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn

Combine all the solutions

x = \frac{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn

Popular Examples

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

cos(x)= 7/25

cos (x) = \frac{7}{25}

2sin(3x)+sqrt(2)=0

2 sin (3 x) + 2 = 0

0=cos(θ)

0 = cos (θ)

sec^2(x)+3tan(x)=5

sec^{2} (x) + 3 tan (x) = 5

Frequently Asked Questions (FAQ)

What is the general solution for 3cos^2(x)-5cos(x)=1+3sin^2(x) ?
The general solution for 3cos^2(x)-5cos(x)=1+3sin^2(x) is x=(2pi)/3+2pin,x=(4pi)/3+2pin