What is the general solution for 4cos(x)+sin^2(x)=3cos^2(x)+2 ?

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

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4cos(x)+sin^2(x)=3cos^2(x)+2

Solution

4 cos (x) + sin^{2} (x) = 3 cos^{2} (x) + 2

Solution

x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

Degrees

x = 6 0^{\circ} + 36 0^{\circ} n, x = 30 0^{\circ} + 36 0^{\circ} n

Solution steps

4 cos (x) + sin^{2} (x) = 3 cos^{2} (x) + 2

Subtract

3 cos^{2} (x) + 2

from both sides

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

Rewrite using trig identities

- 2 + sin^{2} (x) - 3 cos^{2} (x) + 4 cos (x)

Use the Pythagorean identity:

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

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

= - 2 + 1 - cos^{2} (x) - 3 cos^{2} (x) + 4 cos (x)

Simplify

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

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

Add similar elements:

- cos^{2} (x) - 3 cos^{2} (x) = - 4 cos^{2} (x)

= - 2 + 1 - 4 cos^{2} (x) + 4 cos (x)

Add/Subtract the numbers:

- 2 + 1 = - 1

= 4 cos (x) - 4 cos^{2} (x) - 1

= 4 cos (x) - 4 cos^{2} (x) - 1

- 1 + 4 cos (x) - 4 cos^{2} (x) = 0

Solve by substitution

- 1 + 4 cos (x) - 4 cos^{2} (x) = 0

Let:

cos (x) = u

- 1 + 4 u - 4 u^{2} = 0

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

- 1 + 4 u - 4 u^{2} = 0

Write in the standard form

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

- 4 u^{2} + 4 u - 1 = 0

Solve with the quadratic formula

- 4 u^{2} + 4 u - 1 = 0

Quadratic Equation Formula:

For

a = - 4, b = 4, c = - 1

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

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

4^{2} - 4 (- 4) (- 1) = 0

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

Apply rule

- (- a) = a

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

Multiply the numbers:

4 \cdot 4 \cdot 1 = 16

= 4^{2} - 16

4^{2} = 16

= 16 - 16

Subtract the numbers:

16 - 16 = 0

= 0

u_{1, 2} = \frac{- 4 \pm 0}{2 ( - 4 )}

u = \frac{- 4}{2 ( - 4 )}

\frac{- 4}{2 ( - 4 )} = \frac{1}{2}

\frac{- 4}{2 ( - 4 )}

Remove parentheses:

(- a) = - a

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

Multiply the numbers:

2 \cdot 4 = 8

= \frac{- 4}{- 8}

Apply the fraction rule:

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

= \frac{4}{8}

Cancel the common factor:

4

= \frac{1}{2}

u = \frac{1}{2}

The solution to the quadratic equation is:

u = \frac{1}{2}

Substitute back

u = cos (x)

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

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

cos (x) = \frac{1}{2} : x = \frac{π}{3} + 2 πn, x = \frac{5 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{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

Combine all the solutions

x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

Popular Examples

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5 tan (B) + 13 = 0

tan^2(x)=sqrt(3)tan(x)

tan^{2} (x) = 3 tan (x)

2cos^2(x)-2=0

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

sin(θ-pi/4)-1=0

sin (θ - \frac{π}{4}) - 1 = 0

tan^2(x)+csc^2(x)-3=0

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

Frequently Asked Questions (FAQ)

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