What is the general solution for sin(t)= 3/5 ,cos(t)= 4/5 ?

The general solution for sin(t)= 3/5 ,cos(t)= 4/5 is No Solution for t\in\mathbb{R}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

sin(t)= 3/5 ,cos(t)= 4/5

sin (t) = \frac{3}{5}, cos (t) = \frac{4}{5}

No Solution for t \in R

Solution steps

sin (t) = \frac{3}{5}, cos (t) = \frac{4}{5}

Apply trig inverse properties

t = arcsin (\frac{3}{5}) + 2 πn, t = π - arcsin (\frac{3}{5}) + 2 πn

Solutions for the range

cos (t) = \frac{4}{5}

No Solution for t \in R

cos (3 x) = 4 cos^{2} (x) - 3 cos (x)

2 sin (2 θ) = 0

- sin (2 x) = - 2 sin (x)

sin (8 x + 2) = cos (6 x - 10)

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

What is the general solution for sin(t)= 3/5 ,cos(t)= 4/5 ?
The general solution for sin(t)= 3/5 ,cos(t)= 4/5 is No Solution for t\in\mathbb{R}