What are the solutions of 6 tanx sinx-6 tanx=0 in the interval [0,2pi]?

Question

What are the solutions of 6 tanx sinx-6 tanx=0 in the interval ​[0,2pi]?

Delilah Aguilar · Accepted Answer

#x=0, pi, 2pi# Factor out the #6tanx#: #6tanx(sinx-1)=0# #tanx=0# #x=pi, 0, 2pi# #sinx=1# #x=pi/2# Unfortunately tangent is undefined at #pi/2# so feasible solutions: #x=0, pi, 2pi#

Aaliyah Abdullah · Answer

#0; pi/2; pi; 2pi# 6tan x.sin x - 6tan x = 0 6tan x(sin x - 1) = 0 Either factor should be zero. a. sin x - 1 = 0 --> sin x = 1 Unit circle gives --> #x = pi/2# b. tan x = 0 Unit circle --> x = 0 and #x = pi, and x = 2pi# Answers for #[0, 2pi]#: #0, pi/2, pi, 2pi#

Zoe Austin · Answer

undefined To solve the equation $6 \tan x \sin x - 6 \tan x = 0$ in the interval $[0, 2\pi]$, we can first factor out $6 \tan x$ to get:

$$6 \tan x (\sin x - 1) = 0.$$

This equation is satisfied if either $6 \tan x = 0$ or $\sin x - 1 = 0$.

1. **$6 \tan x = 0$:** This occurs when $\tan x = 0$. In the interval $[0, 2\pi]$, the solutions for $\tan x = 0$ are $x = 0, \pi, 2\pi$.

2. **$\sin x - 1 = 0$:** This occurs when $\sin x = 1$. In the interval $[0, 2\pi]$, the solution for $\sin x = 1$ is $x = \frac{\pi}{2}$.

Therefore, the solutions of the equation $6 \tan x \sin x - 6 \tan x = 0$ in the interval $[0, 2\pi]$ are $x = 0, \frac{\pi}{2}, \pi, 2\pi$.

What are the solutions of 6 tanx sinx-6 tanx=0 in the interval ​[0,2pi]?

What are the solutions of 6 tanx sinx-6 tanx=0 in the interval [0,2pi]?