Question

Solve the equation on the interval [0, 2π).

tan2x sin x = tan2x

A. 0, (pi)/2

B. (pi)/2 , 2π

C. (pi)/2 , π

D. 0, π

Answer 1

Answer: The answer is (A)
`0,~(\pi)/(2).`

Step-by-step explanation: The given trigonometric equation is

`\tan 2x\sin x=\tan 2x.`

We are to solve the above equation in the interval
`[0,2\pi).`

The solution is as follows:

`\tan 2x\sin x=\tan 2x\\\\\Rightarrow \tan2x\sin x-\tan 2x=0\\\\\Rightarrow \tan 2x(\sin x-1)=0\\\\\Rightarrow \tan 2x=0,~~~~~~~\sin x-1=0\\\\\Rightarrow \tan 2x=\tan 0,~\Rightarrow \sin x=1\\\\\Rightarrow 2x=0,~~~~~~~\Rightarrow \sin x=\sin (\pi)/(2)\\\\\\\Rightarrow x=0,~~~~~~~~~~\Rightarrow x=(\pi)/(2).`

Thus, the correct option is (A).

Answer 2

tan2x sin x = tan2x
tan2xsin x - tan 2x = 0
tan 2x(sinx - 1) = 0
tan 2x = 0
or sin x = 1
so one root is pi and other is 0
so correct option is A
hope it helps