Question

How do you solve $\sin (3x)=-1$ with domain between $0$ to $2\pi $ ?

Answer 1

Hint: In this question, we have to find the value of x. The equation given to us is in the form of trigonometric functions, so we will apply the trigonometric formulas to get the required solution to the problem. We first take ${{\sin }^{-1}}$ on both sides of the equation and then apply the formulas ${{\sin }^{-1}}\left( \sin x \right)=x$ on the left-hand sides of the equation. After the necessary calculations, we will write ${{\sin }^{-1}}(-1)=\dfrac{3\pi }{2}+2n\pi $ , where n is a positive integer. Also, we add $2n\pi $ because the sin function is a periodic function. Then, on further simplification, we get three different values of x, which is our required solution.

Complete step-by-step answer:
According to the question, it is given that there is an equation and we have to solve for x.
So, we will apply the trigonometric formula.
The equation given to us is $\sin (3x)=-1$ -------- (1)
Now, we first take ${{\sin }^{-1}}$ on both sides in the equation (1), we get
${{\sin }^{-1}}\left( \sin (3x) \right)={{\sin }^{-1}}\left( -1 \right)$
Now, we will apply the trigonometric formula ${{\sin }^{-1}}\left( \sin x \right)=x$ on the left-hand sides of the above equation, we get
$3x={{\sin }^{-1}}(-1)$ ---------- (2)
As we know, the sin function is a periodic function, therefore its value at the angle (-1), is equal to
${{\sin }^{-1}}(-1)=\dfrac{3\pi }{2}+2n\pi $ , where n is a positive integer -------- (3)
Therefore, putting the value of equation (3) in equation (2), we get
$3x=\dfrac{3\pi }{2}+2n\pi $
Now, we will divide 3 on both sides of the equation, we get
$\dfrac{3}{3}x=\dfrac{\dfrac{3\pi }{2}+2n\pi }{3}$
On further simplification, we get
$x=\dfrac{\pi }{2}+\dfrac{2}{3}n\pi $ ------------- (4)
Now, we will put three different values of n, that is 0, 1, and 2 because the range of x is 0 to $2\pi $, therefore we get
$x=\dfrac{\pi }{2}+\dfrac{2}{3}.(0)\pi $ , ------- (5)
$x=\dfrac{\pi }{2}+\dfrac{2}{3}.(1)\pi $ , and --------- (6)
$x=\dfrac{\pi }{2}+\dfrac{2}{3}.(2).\pi $ --------- (7)
So, we will solve equation (5), we get
$x=\dfrac{\pi }{2}$ ,
So, we will solve equation (6), we get
$x=\dfrac{\pi }{2}+\dfrac{2}{3}.(1)\pi $
Therefore, we get
$x=\dfrac{\pi }{2}+\dfrac{2\pi }{3}$
So, on taking LCM of the above equation, we get
$x=\dfrac{7\pi }{6}$
So, we will solve equation (7), we get
$x=\dfrac{\pi }{2}+\dfrac{2}{3}.(2).\pi $
Therefore, we get
$x=\dfrac{\pi }{2}+\dfrac{4\pi }{3}$
So, on taking LCM of the above equation, we get
$x=\dfrac{11\pi }{6}$
Therefore, for the equation $\sin (3x)=-1$ with the domain between $0$ to $2\pi $ , the value of x is $\dfrac{\pi }{2},\dfrac{7\pi }{6},\dfrac{11\pi }{6}$

Note: While solving this problem, do mention all the trigonometric formulas you are using to avoid mathematical errors. Do not forget about the range of x, that is we have to take 3 different values of x because ${{\sin }^{-1}}(-1)=\dfrac{3\pi }{2}+2n\pi $, so in the given range it will take 3 angles with the same value, that is -1.