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

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

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Popular >

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

Solution

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

Solution

x = 2 πn

Degrees

x = 0^{\circ} + 36 0^{\circ} n

Solution steps

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

Subtract

- sin^{2} (x) + 4

from both sides

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

Simplify

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

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

Solve by substitution

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

Let:

cos (x) = u

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

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

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

Write in the standard form

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

- u^{2} + 4 u - 3 = 0

Solve with the quadratic formula

- u^{2} + 4 u - 3 = 0

Quadratic Equation Formula:

For

a = - 1, b = 4, c = - 3

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

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

4^{2} - 4 (- 1) (- 3) = 2

4^{2} - 4 (- 1) (- 3)

Apply rule

- (- a) = a

= 4^{2} - 4 \cdot 1 \cdot 3

Multiply the numbers:

4 \cdot 1 \cdot 3 = 12

= 4^{2} - 12

4^{2} = 16

= 16 - 12

Subtract the numbers:

16 - 12 = 4

= 4

Factor the number:

4 = 2^{2}

= 2^{2}

Apply radical rule:

2^{2} = 2

= 2

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

Separate the solutions

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

u = \frac{- 4 + 2}{2 ( - 1 )} : 1

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

Remove parentheses:

(- a) = - a

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

Add/Subtract the numbers:

- 4 + 2 = - 2

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

Multiply the numbers:

2 \cdot 1 = 2

= \frac{- 2}{- 2}

Apply the fraction rule:

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

= \frac{2}{2}

Apply rule

\frac{a}{a} = 1

= 1

u = \frac{- 4 - 2}{2 ( - 1 )} : 3

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

Remove parentheses:

(- a) = - a

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

Subtract the numbers:

- 4 - 2 = - 6

= \frac{- 6}{- 2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{- 6}{- 2}

Apply the fraction rule:

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

= \frac{6}{2}

Divide the numbers:

\frac{6}{2} = 3

= 3

The solutions to the quadratic equation are:

u = 1, u = 3

Substitute back

u = cos (x)

cos (x) = 1, cos (x) = 3

cos (x) = 1, cos (x) = 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) = 3 :

No Solution

cos (x) = 3

- 1 \leq cos (x) \leq 1

No Solution

Combine all the solutions

x = 2 πn

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8tan(b)+12=3tan(b)+7

8 tan (b) + 12 = 3 tan (b) + 7

Frequently Asked Questions (FAQ)

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