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

The general solution for 5cos^2(x)-2cos(x)-3=sin^2(x) is x=2pin,x=2.30052…+2pin,x=-2.30052…+2pin

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

Solution

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

Solution

x = 2 πn, x = 2.30052 \dots + 2 πn, x = - 2.30052 \dots + 2 πn

Degrees

x = 0^{\circ} + 36 0^{\circ} n, x = 131.81031 \dots^{\circ} + 36 0^{\circ} n, x = - 131.81031 \dots^{\circ} + 36 0^{\circ} n

Solution steps

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

Subtract

sin^{2} (x)

from both sides

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

Simplify

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

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

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

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

Distribute parentheses

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

Apply minus-plus rules

- (- a) = a, - (a) = - a

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

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

Simplify

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

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

Group like terms

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

Add similar elements:

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

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

Subtract the numbers:

- 3 - 1 = - 4

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

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

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

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

Solve by substitution

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

Let:

cos (x) = u

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

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

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

Write in the standard form

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

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

a = 6, b = - 2, c = - 4

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

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

(- 2)^{2} - 4 \cdot 6 (- 4) = 10

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

Apply rule

- (- a) = a

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

Apply exponent rule:

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

n

is even

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

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

Multiply the numbers:

4 \cdot 6 \cdot 4 = 96

= 2^{2} + 96

2^{2} = 4

= 4 + 96

Add the numbers:

4 + 96 = 100

= 100

Factor the number:

100 = 1 0^{2}

= 1 0^{2}

Apply radical rule:

1 0^{2} = 10

= 10

u_{1, 2} = \frac{- ( - 2 ) \pm 10}{2 \cdot 6}

Separate the solutions

u_{1} = \frac{- ( - 2 ) + 10}{2 \cdot 6}, u_{2} = \frac{- ( - 2 ) - 10}{2 \cdot 6}

u = \frac{- ( - 2 ) + 10}{2 \cdot 6} : 1

\frac{- ( - 2 ) + 10}{2 \cdot 6}

Apply rule

- (- a) = a

= \frac{2 + 10}{2 \cdot 6}

Add the numbers:

2 + 10 = 12

= \frac{12}{2 \cdot 6}

Multiply the numbers:

2 \cdot 6 = 12

= \frac{12}{12}

Apply rule

\frac{a}{a} = 1

= 1

u = \frac{- ( - 2 ) - 10}{2 \cdot 6} : - \frac{2}{3}

\frac{- ( - 2 ) - 10}{2 \cdot 6}

Apply rule

- (- a) = a

= \frac{2 - 10}{2 \cdot 6}

Subtract the numbers:

2 - 10 = - 8

= \frac{- 8}{2 \cdot 6}

Multiply the numbers:

2 \cdot 6 = 12

= \frac{- 8}{12}

Apply the fraction rule:

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

= - \frac{8}{12}

Cancel the common factor:

4

= - \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 = - 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 = - arccos (- a) + 2 πn

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

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

Combine all the solutions

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

Show solutions in decimal form

x = 2 πn, x = 2.30052 \dots + 2 πn, x = - 2.30052 \dots + 2 πn

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Frequently Asked Questions (FAQ)

What is the general solution for 5cos^2(x)-2cos(x)-3=sin^2(x) ?
The general solution for 5cos^2(x)-2cos(x)-3=sin^2(x) is x=2pin,x=2.30052…+2pin,x=-2.30052…+2pin