What is the general solution for -4cos(x)=-sin^2(x)+4,0<= x<= 2pi ?

Q: What is the general solution for -4cos(x)=-sin^2(x)+4,0<= x<= 2pi ?

The general solution for -4cos(x)=-sin^2(x)+4,0<= x<= 2pi is x=pi

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Popular >

-4cos(x)=-sin^2(x)+4,0<= x<= 2pi

Solution

- 4 cos (x) = - sin^{2} (x) + 4, 0 \leq x \leq 2 π

Solution

x = π

Degrees

x = 18 0^{\circ}

Solution steps

- 4 cos (x) = - sin^{2} (x) + 4, 0 \leq x \leq 2 π

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

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

- 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

Apply exponent rule:

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

n

is even

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

= 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 )} : - 3

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

Remove parentheses:

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

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

Add 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

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

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

Remove parentheses:

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

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

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

The solutions to the quadratic equation are:

u = - 3, u = - 1

Substitute back

u = cos (x)

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

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

cos (x) = - 3, 0 \leq x \leq 2 π :

No Solution

cos (x) = - 3, 0 \leq x \leq 2 π

- 1 \leq cos (x) \leq 1

No Solution

cos (x) = - 1, 0 \leq x \leq 2 π : x = π

cos (x) = - 1, 0 \leq x \leq 2 π

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 = π + 2 πn

x = π + 2 πn

Solutions for the range

0 \leq x \leq 2 π

x = π

Combine all the solutions

x = π

Popular Examples

0=cos^2(pi/(12)t)

0 = cos^{2} (\frac{π}{12} t)

sec^2(x)+2sec(x)+4=0

sec^{2} (x) + 2 sec (x) + 4 = 0

cos(A)=0.7865

cos (A) = 0.7865

sin(x)=cos(2x+51)

sin (x) = cos (2 x + 5 1^{\circ})

sin(x)= pi/4

sin (x) = \frac{π}{4}

Frequently Asked Questions (FAQ)

What is the general solution for -4cos(x)=-sin^2(x)+4,0<= x<= 2pi ?
The general solution for -4cos(x)=-sin^2(x)+4,0<= x<= 2pi is x=pi