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

The general solution for sin(t)cos(t)=sin^2(t) is t=2pin,t=pi+2pin,t= pi/4+pin

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

sin(t)cos(t)=sin^2(t)

Solution

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

Solution

t = 2 πn, t = π + 2 πn, t = \frac{π}{4} + πn

Degrees

t = 0^{\circ} + 36 0^{\circ} n, t = 18 0^{\circ} + 36 0^{\circ} n, t = 4 5^{\circ} + 18 0^{\circ} n

Solution steps

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

Subtract

sin^{2} (t)

from both sides

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

Factor

sin (t) cos (t) - sin^{2} (t) : sin (t) (cos (t) - sin (t))

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

Apply exponent rule:

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

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

= sin (t) cos (t) - sin (t) sin (t)

Factor out common term

sin (t)

= sin (t) (cos (t) - sin (t))

sin (t) (cos (t) - sin (t)) = 0

Solving each part separately

sin (t) = 0 or cos (t) - sin (t) = 0

sin (t) = 0 : t = 2 πn, t = π + 2 πn

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 (t) - sin (t) = 0 : t = \frac{π}{4} + πn

cos (t) - sin (t) = 0

Rewrite using trig identities

cos (t) - sin (t) = 0

Divide both sides by

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

\frac{cos ( t ) - sin ( t )}{cos ( t )} = \frac{0}{cos ( t )}

Simplify

1 - \frac{sin ( t )}{cos ( t )} = 0

Use the basic trigonometric identity:

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

1 - tan (t) = 0

1 - tan (t) = 0

Move

1

to the right side

1 - tan (t) = 0

Subtract

1

from both sides

1 - tan (t) - 1 = 0 - 1

Simplify

- tan (t) = - 1

- tan (t) = - 1

Divide both sides by

- 1

- tan (t) = - 1

Divide both sides by

- 1

\frac{- tan ( t )}{- 1} = \frac{- 1}{- 1}

Simplify

tan (t) = 1

tan (t) = 1

General solutions for

tan (t) = 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}

t = \frac{π}{4} + πn

t = \frac{π}{4} + πn

Combine all the solutions

t = 2 πn, t = π + 2 πn, t = \frac{π}{4} + πn

Popular Examples

cos(x)=(290)/(346.7)

cos (x) = \frac{290}{346.7}

tan(x)= 19/11

tan (x) = \frac{19}{11}

cot(x)=cot(3x-50),0<x<180

cot (x) = cot (3 x - 5 0^{\circ}), 0 < x < 18 0^{\circ}

tan(2θ)= 4/3

tan (2 θ) = \frac{4}{3}

3sec^2(x)+2-8tan(x)=0

3 sec^{2} (x) + 2 - 8 tan (x) = 0

Frequently Asked Questions (FAQ)

What is the general solution for sin(t)cos(t)=sin^2(t) ?
The general solution for sin(t)cos(t)=sin^2(t) is t=2pin,t=pi+2pin,t= pi/4+pin