Question

Choose the values that would complete the general solution for y = arcsin(-sqrt3/2). ___±2k π

Answer 1

Answer:

The required term is $y=-\frac{\pi}{3}\pm2k\pi$

Step-by-step explanation:

Given : $y=\sin^{-1}(-\frac{\sqrt3}{2} )$

To find : The general solution ?

Solution :

$y=\sin^{-1}(-\frac{\sqrt3}{2})$

Can be re-written as

$\sin y=-\frac{\sqrt3}{2}$

We know, $\sin (\frac{\pi}{3})=\frac{\sqrt3}{2}$

$\sin (y)=\sin (-\frac{\pi}{3})$

$y=-\frac{\pi}{3}$

The general solution of the sine function is

$y=-\frac{\pi}{3}\pm2k\pi$

Therefore, The required term is $y=-\frac{\pi}{3}\pm2k\pi$      

Answer 2

To solve this problem you must apply the proccedure shown below:
 1. You have the following inverse trigonometric function given in the problem above:
 $y=arcsin(- \sqrt{3} /2)$
 2. By definition, you have that $y=arcsinx$ is equivalent to $siny=x$ and $sin(-x)=-sin(x)$, therefore:
 $-siny=( \sqrt{3}/2)$
 4. The value of the sine of $\sqrt{3}/2$ is $\pi/3$, therefore:
 $y=- \pi/3$
 The answer is:$- \pi/3$