What is the general solution for sin^3(x)+cos^3(x)=(1-1)/(2sin^2(x)) ?

The general solution for sin^3(x)+cos^3(x)=(1-1)/(2sin^2(x)) is x=(3pi)/4+pin

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

sin^3(x)+cos^3(x)=(1-1)/(2sin^2(x))

Solution

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

Solution

x = \frac{3 π}{4} + πn

Degrees

x = 13 5^{\circ} + 18 0^{\circ} n

Solution steps

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

Subtract

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

from both sides

sin^{3} (x) + cos^{3} (x) = 0

Factor

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

sin^{3} (x) + cos^{3} (x)

Apply Sum of Cubes Formula:

x^{3} + y^{3} = (x + y) (x^{2} - x y + y^{2})

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

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

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

= (cos (x) + sin (x)) (- cos (x) sin (x) + 1)

(cos (x) + sin (x)) (- cos (x) sin (x) + 1) = 0

Solving each part separately

cos (x) + sin (x) = 0 or - cos (x) sin (x) + 1 = 0

cos (x) + sin (x) = 0 : x = \frac{3 π}{4} + πn

cos (x) + sin (x) = 0

Rewrite using trig identities

cos (x) + sin (x) = 0

Divide both sides by

cos (x), cos (x) \neq = 0

\frac{cos ( x ) + sin ( x )}{cos ( x )} = \frac{0}{cos ( x )}

Simplify

1 + \frac{sin ( x )}{cos ( x )} = 0

Use the basic trigonometric identity:

\frac{sin ( x )}{cos ( x )} = tan (x)

1 + tan (x) = 0

1 + tan (x) = 0

Move

1

to the right side

1 + tan (x) = 0

Subtract

1

from both sides

1 + tan (x) - 1 = 0 - 1

Simplify

tan (x) = - 1

tan (x) = - 1

General solutions for

tan (x) = - 1

tan (x)

periodicity table with

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} tan (x) 0 \frac{3}{3} 13 \pm \infty - 3 - 1 - \frac{3}{3}

x = \frac{3 π}{4} + πn

x = \frac{3 π}{4} + πn

- cos (x) sin (x) + 1 = 0 :

No Solution

- cos (x) sin (x) + 1 = 0

Rewrite using trig identities

- cos (x) sin (x) + 1

Use the Double Angle identity:

2 sin (x) cos (x) = sin (2 x)

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

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

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

Move

1

to the right side

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

Subtract

1

from both sides

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

Simplify

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

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

Multiply both sides by

2

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

Multiply both sides by

2

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

Simplify

- sin (2 x) = - 2

- sin (2 x) = - 2

Divide both sides by

- 1

- sin (2 x) = - 2

Divide both sides by

- 1

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

Simplify

sin (2 x) = 2

sin (2 x) = 2

- 1 \leq sin (x) \leq 1

No Solution

Combine all the solutions

x = \frac{3 π}{4} + πn

Popular Examples

solvefor i,xsin^2(x)=cos^2(x)solve for

i, x sin^{2} (x) = cos^{2} (x)

solvefor c,sin^2(x)+12=12(sin(x)-cos(x))solve for

c, sin^{2} (x) + 12 = 12 (sin (x) - cos (x))

solvefor c,sin(x)-5cos(x)=sin^2(x)+5cos^2(x)solve for

c, sin (x) - 5 cos (x) = sin^{2} (x) + 5 cos^{2} (x)

solvefor i,sin(x)+sin^2(x)+cos^3(x)=0 solve for

i, sin (x) + sin^{2} (x) + cos^{3} (x) = 0

y=2sin(cos(+0))

y = 2 sin (cos (+ 0))

Frequently Asked Questions (FAQ)

What is the general solution for sin^3(x)+cos^3(x)=(1-1)/(2sin^2(x)) ?
The general solution for sin^3(x)+cos^3(x)=(1-1)/(2sin^2(x)) is x=(3pi)/4+pin