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

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

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sin^2(t)sin(t)=0

sin^{2} (t) sin (t) = 0

t = 2 πn, t = π + 2 πn

Degrees

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

Solution steps

sin^{2} (t) sin (t) = 0

Simplify

sin^{2} (t) sin (t) : sin^{3} (t)

sin^{2} (t) sin (t)

Apply exponent rule:

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

sin^{2} (t) sin (t) = sin^{2 + 1} (t)

= sin^{2 + 1} (t)

Add the numbers:

2 + 1 = 3

= sin^{3} (t)

sin^{3} (t) = 0

Apply rule

x^{n} = 0 \Rightarrow x = 0

sin (t) = 0

General solutions for

sin (t) = 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}

t = 0 + 2 πn, t = π + 2 πn

t = 0 + 2 πn, t = π + 2 πn

Solve

t = 0 + 2 πn : t = 2 πn

t = 0 + 2 πn

0 + 2 πn = 2 πn

t = 2 πn

t = 2 πn, t = π + 2 πn

cos (x) = \frac{100}{130}

sec^{2} (θ) - 1 = 0, 0 \leq θ \leq 2 π

2 cos^{2} (θ) - 3 cos (θ) = 0, \frac{3 π}{2} < θ < 2 π

cos (θ) = \frac{9}{17}

- 1 = \frac{tan ( x ) - ( \frac{3}{4} )}{1 + tan ( x ) ( \frac{3}{4} )}

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