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

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

The general solution for -4sin(x)=-cos^2(x)+4,0<= x<2pi is x=(3pi)/2

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Popular >

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

Solution

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

Solution

x = \frac{3 π}{2}

Degrees

x = 27 0^{\circ}

Solution steps

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

Subtract

- cos^{2} (x) + 4

from both sides

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

Simplify

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

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

Solve by substitution

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

Let:

sin (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 = sin (x)

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

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

sin (x) = - 3, 0 \leq x < 2 π :

No Solution

sin (x) = - 3, 0 \leq x < 2 π

- 1 \leq sin (x) \leq 1

No Solution

sin (x) = - 1, 0 \leq x < 2 π : x = \frac{3 π}{2}

sin (x) = - 1, 0 \leq x < 2 π

General solutions for

sin (x) = - 1

sin (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} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{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} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

x = \frac{3 π}{2} + 2 πn

x = \frac{3 π}{2} + 2 πn

Solutions for the range

0 \leq x < 2 π

x = \frac{3 π}{2}

Combine all the solutions

x = \frac{3 π}{2}

Popular Examples

cos(x)=(sqrt(3))/(sqrt(2))

cos (x) = \frac{3}{2}

sin(θ)=-(sqrt(3))/2 ,0<= θ<2pi

sin (θ) = - \frac{3}{2}, 0 \leq θ < 2 π

sin(x)=7938894612

sin (x) = 7938894612

3sin(2θ)=5cos(θ)

3 sin (2 θ) = 5 cos (θ)

cos(2x)= 1/2 ,(0,2pi)

cos (2 x) = \frac{1}{2}, (0, 2 π)

Frequently Asked Questions (FAQ)

What is the general solution for -4sin(x)=-cos^2(x)+4,0<= x<2pi ?
The general solution for -4sin(x)=-cos^2(x)+4,0<= x<2pi is x=(3pi)/2