What is the general solution for -4sin(x)=cos^2(x)+1,0<= x<= 2pi ?

Q: What is the general solution for -4sin(x)=cos^2(x)+1,0<= x<= 2pi ?

The general solution for -4sin(x)=cos^2(x)+1,0<= x<= 2pi is x=pi+0.46619…,x=-0.46619…+2pi

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

-4sin(x)=cos^2(x)+1,0<= x<= 2pi

Solution

- 4 sin (x) = cos^{2} (x) + 1, 0 \leq x \leq 2 π

Solution

x = π + 0.46619 \dots, x = - 0.46619 \dots + 2 π

Degrees

x = 206.71095 \dots^{\circ}, x = 333.28904 \dots^{\circ}

Solution steps

- 4 sin (x) = cos^{2} (x) + 1, 0 \leq x \leq 2 π

Subtract

cos^{2} (x) + 1

from both sides

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

Simplify

- 1 - (1 - sin^{2} (x)) - 4 sin (x) : sin^{2} (x) - 4 sin (x) - 2

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

- (1 - sin^{2} (x)) : - 1 + sin^{2} (x)

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

Distribute parentheses

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

Apply minus-plus rules

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

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

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

Subtract the numbers:

- 1 - 1 = - 2

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

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

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

Solve by substitution

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

Let:

sin (x) = u

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

- 2 + u^{2} - 4 u = 0 : u = 2 + 6, u = 2 - 6

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

Write in the standard form

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

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

a = 1, b = - 4, c = - 2

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

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

(- 4)^{2} - 4 \cdot 1 \cdot (- 2) = 26

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

Apply rule

- (- a) = a

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

Apply exponent rule:

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

n

is even

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

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

Multiply the numbers:

4 \cdot 1 \cdot 2 = 8

= 4^{2} + 8

4^{2} = 16

= 16 + 8

Add the numbers:

16 + 8 = 24

= 24

Prime factorization of

24 : 2^{3} \cdot 3

24

24

divides by

224 = 12 \cdot 2

= 2 \cdot 12

12

divides by

212 = 6 \cdot 2

= 2 \cdot 2 \cdot 6

6

divides by

26 = 3 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 3

2, 3

are all prime numbers, therefore no further factorization is possible

= 2 \cdot 2 \cdot 2 \cdot 3

= 2^{3} \cdot 3

= 2^{3} \cdot 3

Apply exponent rule:

a^{b + c} = a^{b} \cdot a^{c}

= 2^{2} \cdot 2 \cdot 3

Apply radical rule:

= 2^{2} 2 \cdot 3

Apply radical rule:

2^{2} = 2

= 2 2 \cdot 3

Refine

= 26

u_{1, 2} = \frac{- ( - 4 ) \pm 2 6}{2 \cdot 1}

Separate the solutions

u_{1} = \frac{- ( - 4 ) + 2 6}{2 \cdot 1}, u_{2} = \frac{- ( - 4 ) - 2 6}{2 \cdot 1}

u = \frac{- ( - 4 ) + 2 6}{2 \cdot 1} : 2 + 6

\frac{- ( - 4 ) + 2 6}{2 \cdot 1}

Apply rule

- (- a) = a

= \frac{4 + 2 6}{2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{4 + 2 6}{2}

Factor

4 + 26 : 2 (2 + 6)

4 + 26

Rewrite as

= 2 \cdot 2 + 26

Factor out common term

2

= 2 (2 + 6)

= \frac{2 ( 2 + 6 )}{2}

Divide the numbers:

\frac{2}{2} = 1

= 2 + 6

u = \frac{- ( - 4 ) - 2 6}{2 \cdot 1} : 2 - 6

\frac{- ( - 4 ) - 2 6}{2 \cdot 1}

Apply rule

- (- a) = a

= \frac{4 - 2 6}{2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{4 - 2 6}{2}

Factor

4 - 26 : 2 (2 - 6)

4 - 26

Rewrite as

= 2 \cdot 2 - 26

Factor out common term

2

= 2 (2 - 6)

= \frac{2 ( 2 - 6 )}{2}

Divide the numbers:

\frac{2}{2} = 1

= 2 - 6

The solutions to the quadratic equation are:

u = 2 + 6, u = 2 - 6

Substitute back

u = sin (x)

sin (x) = 2 + 6, sin (x) = 2 - 6

sin (x) = 2 + 6, sin (x) = 2 - 6

sin (x) = 2 + 6, 0 \leq x \leq 2 π :

No Solution

sin (x) = 2 + 6, 0 \leq x \leq 2 π

- 1 \leq sin (x) \leq 1

No Solution

sin (x) = 2 - 6, 0 \leq x \leq 2 π : x = π + arcsin (6 - 2), x = arcsin (2 - 6) + 2 π

sin (x) = 2 - 6, 0 \leq x \leq 2 π

Apply trig inverse properties

sin (x) = 2 - 6

General solutions for

sin (x) = 2 - 6

sin (x) = - a \Rightarrow x = arcsin (- a) + 2 πn, x = π + arcsin (a) + 2 πn

x = arcsin (2 - 6) + 2 πn, x = π + arcsin (- 2 + 6) + 2 πn

x = arcsin (2 - 6) + 2 πn, x = π + arcsin (- 2 + 6) + 2 πn

Solutions for the range

0 \leq x \leq 2 π

x = π + arcsin (6 - 2), x = arcsin (2 - 6) + 2 π

Combine all the solutions

x = π + arcsin (6 - 2), x = arcsin (2 - 6) + 2 π

Show solutions in decimal form

x = π + 0.46619 \dots, x = - 0.46619 \dots + 2 π

Popular Examples

sqrt(3)sin(x)-3cos(x)=sqrt(3)

3 sin (x) - 3 cos (x) = 3

tan(2x)cos(2x)+cot(2x)sin(2x)=1

tan (2 x) cos (2 x) + cot (2 x) sin (2 x) = 1

sin^2(2x)-cos^2(2x)= 1/2

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

sin(x)=(1.2)/(sqrt(10))

sin (x) = \frac{1.2}{10}

solvefor t,x=-3cos(pit)solve for

t, x = - 3 cos (π t)

Frequently Asked Questions (FAQ)

What is the general solution for -4sin(x)=cos^2(x)+1,0<= x<= 2pi ?
The general solution for -4sin(x)=cos^2(x)+1,0<= x<= 2pi is x=pi+0.46619…,x=-0.46619…+2pi