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

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

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

Solution

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

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

Subtract

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

from both sides

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

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

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

Subtract the numbers:

2 - 1 = 1

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

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

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

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

4^{2} - 4 \cdot 4 \cdot 1 = 0

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 \cdot 4}

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

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

\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{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

solvefor x,ln(y)=arctan(x)+C solve for

x, ln (y) = arctan (x) + C

sin(x)+2sin(x)=5

sin (x) + 2 sin (x) = 5

cos(θ)=-3/10

cos (θ) = - \frac{3}{10}

2cos^3(x)=0

2 cos^{3} (x) = 0

50sin(x)+15cos(x)=40,0<x<pi

50 sin (x) + 15 cos (x) = 40, 0 < x < π

Frequently Asked Questions (FAQ)

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