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

The general solution for 2cot^2(t)sin(t)-cot^2(t)=0 is t= pi/2+pin,t= pi/6+2pin,t=(5pi)/6+2pin

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

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

t = \frac{π}{2} + πn, t = \frac{π}{6} + 2 πn, t = \frac{5 π}{6} + 2 πn

Degrees

t = 9 0^{\circ} + 18 0^{\circ} n, t = 3 0^{\circ} + 36 0^{\circ} n, t = 15 0^{\circ} + 36 0^{\circ} n

Solution steps

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

Factor

2 cot^{2} (t) sin (t) - cot^{2} (t) : cot^{2} (t) (2 sin (t) - 1)

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

Factor out common term

cot^{2} (t)

= cot^{2} (t) (2 sin (t) - 1)

cot^{2} (t) (2 sin (t) - 1) = 0

Solving each part separately

cot^{2} (t) = 0 or 2 sin (t) - 1 = 0

cot^{2} (t) = 0 : t = \frac{π}{2} + πn

cot^{2} (t) = 0

Apply rule

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

cot (t) = 0

General solutions for

cot (t) = 0

cot (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} cot (x) \mp \infty 31 \frac{3}{3} 0 - \frac{3}{3} - 1 - 3

t = \frac{π}{2} + πn

t = \frac{π}{2} + πn

2 sin (t) - 1 = 0 : t = \frac{π}{6} + 2 πn, t = \frac{5 π}{6} + 2 πn

2 sin (t) - 1 = 0

Move

1

to the right side

2 sin (t) - 1 = 0

Add

1

to both sides

2 sin (t) - 1 + 1 = 0 + 1

Simplify

2 sin (t) = 1

2 sin (t) = 1

Divide both sides by

2

2 sin (t) = 1

Divide both sides by

2

\frac{2 sin ( t )}{2} = \frac{1}{2}

Simplify

sin (t) = \frac{1}{2}

sin (t) = \frac{1}{2}

General solutions for

sin (t) = \frac{1}{2}

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 = \frac{π}{6} + 2 πn, t = \frac{5 π}{6} + 2 πn

t = \frac{π}{6} + 2 πn, t = \frac{5 π}{6} + 2 πn

Combine all the solutions

t = \frac{π}{2} + πn, t = \frac{π}{6} + 2 πn, t = \frac{5 π}{6} + 2 πn

7 cos (x) = - 1

2 sin^{2} (θ) + cos (θ) - 2 = - 1

sin^{2} (x) - 6 cos (x) + 6 = 0

cos (x) = \frac{14.3}{19}

- 12 sin (x) = 0

What is the general solution for 2cot^2(t)sin(t)-cot^2(t)=0 ?
The general solution for 2cot^2(t)sin(t)-cot^2(t)=0 is t= pi/2+pin,t= pi/6+2pin,t=(5pi)/6+2pin