What is the general solution for sin^3(x)sin^2(x)sin(x)=0 ?

The general solution for sin^3(x)sin^2(x)sin(x)=0 is x=2pin,x=pi+2pin

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

sin^3(x)sin^2(x)sin(x)=0

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

x = 2 πn, x = π + 2 πn

Degrees

x = 0^{\circ} + 36 0^{\circ} n, x = 18 0^{\circ} + 36 0^{\circ} n

Solution steps

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

Simplify

sin^{3} (x) sin^{2} (x) sin (x) : sin^{6} (x)

sin^{3} (x) sin^{2} (x) sin (x)

Apply exponent rule:

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

sin^{3} (x) sin^{2} (x) sin (x) = sin^{3 + 2 + 1} (x)

= sin^{3 + 2 + 1} (x)

Add the numbers:

3 + 2 + 1 = 6

= sin^{6} (x)

sin^{6} (x) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

sin (x) = 0

The solution is

sin (x) = 0

General solutions for

sin (x) = 0

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 = 0 + 2 πn, x = π + 2 πn

x = 0 + 2 πn, x = π + 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, x = π + 2 πn

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

sin^{2} (x) cos^{2} (x) = \frac{( 1 - cos ^{4} ( x ) )}{8}

sin^{3} (x) + sin (x) = 2 sin^{22} (x)

\frac{1 + tan ^{2} ( x )}{1 + sec ( x )} = sec (x)

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

What is the general solution for sin^3(x)sin^2(x)sin(x)=0 ?
The general solution for sin^3(x)sin^2(x)sin(x)=0 is x=2pin,x=pi+2pin