en

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

solvefor θ,ma=Tsin(θ)-mg

Solution

solve for

θ, ma = T sin (θ) - m g

Solution

θ = arcsin (\frac{ma + m g}{T}) + 2 πn, θ = π + arcsin (- \frac{ma + m g}{T}) + 2 πn

Solution steps

ma = T sin (θ) - m g

Switch sides

T sin (θ) - m g = ma

Move

m g

to the right side

T sin (θ) - m g = ma

Add

m g

to both sides

T sin (θ) - m g + m g = ma + m g

Simplify

T sin (θ) = ma + m g

T sin (θ) = ma + m g

Divide both sides by

T; T \neq = 0

T sin (θ) = ma + m g

Divide both sides by

T; T \neq = 0

\frac{T sin ( θ )}{T} = \frac{ma}{T} + \frac{m g}{T}; T \neq = 0

Simplify

\frac{T sin ( θ )}{T} = \frac{ma}{T} + \frac{m g}{T}

Simplify

\frac{T sin ( θ )}{T} : sin (θ)

\frac{T sin ( θ )}{T}

Cancel the common factor:

T

= sin (θ)

Simplify

\frac{ma}{T} + \frac{m g}{T} : \frac{ma + m g}{T}

\frac{ma}{T} + \frac{m g}{T}

Apply rule

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

= \frac{ma + m g}{T}

sin (θ) = \frac{ma + m g}{T}; T \neq = 0

sin (θ) = \frac{ma + m g}{T}; T \neq = 0

sin (θ) = \frac{ma + m g}{T}; T \neq = 0

Apply trig inverse properties

sin (θ) = \frac{ma + m g}{T}

General solutions for

sin (θ) = \frac{ma + m g}{T}

sin (x) = a \Rightarrow x = arcsin (a) + 2 πn, x = π + arcsin (a) + 2 πn

θ = arcsin (\frac{ma + m g}{T}) + 2 πn, θ = π + arcsin (- \frac{ma + m g}{T}) + 2 πn

θ = arcsin (\frac{ma + m g}{T}) + 2 πn, θ = π + arcsin (- \frac{ma + m g}{T}) + 2 πn

Popular Examples

sin (x) = \frac{16}{20}

tan(pi/4+x)=2+sqrt(3)

tan (\frac{π}{4} + x) = 2 + 3

3cos(x)=2sin(x)

3 cos (x) = 2 sin (x)

2cos^2(θ)-cos(θ)=1,0<= θ<= 2pi

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

tan(x)-sqrt(3)=0,0<= x<= 2pi

tan (x) - 3 = 0, 0 \leq x \leq 2 π