Question

How do you find the general solutions for $\sin (x+\pi )=0.5$?

Answer 1

In this question, we have to find the general solution f9r the given trigonometric function. We can solve this problem by using the general solution $\sin x=\sin \alpha$ . The general solution of this is $x=n\pi +{{(-1)}^{n}}\alpha$

Complete step by step solution:
In this question we are given a sine function. The given function is $\sin \,(x+\pi )=0.5$
Here, we know that $0.5$ is equal to $\dfrac{1}{2}$
Therefore, the above sine function can be written as below $\sin \,(x+\pi )=\dfrac{1}{2}$ ….. (1)
Now, the sine inverse of $\dfrac{1}{2}$ is equal to $\dfrac{\pi }{6}$ .
${{\sin }^{-1}}\left( \dfrac{1}{2} \right)=\dfrac{\pi }{6}$
Form above equation we conclude that $\sin \left( \dfrac{\pi }{6} \right)=\dfrac{1}{2}$ … (2)
Now, compare equation (1) and equation (2) the value of $\sin (x+\pi )$ and $\sin \left( \dfrac{\pi }{6} \right)$ is equal to $\dfrac{1}{2}$
Therefore, we have
$\sin (x+\pi )=\sin \left( \dfrac{\pi }{6} \right)$
Now, using the result for general solution that is $\sin x=\sin \alpha$ $\Rightarrow$ $x=n\pi +{{(-1)}^{n}}\alpha$
Here $\alpha =\dfrac{\pi }{6}$
Therefore, $x=n\pi +{{(-1)}^{n}}\left( \dfrac{\pi }{6} \right)$ as $x\in z$
Where $z$ is a real number.

Therefore, the required general solution is
$x=n\pi +{{(-1)}^{n}}\left( \dfrac{\pi }{6} \right)$

Note: The general solution is the solution of trigonometric equation which are generalized by using its periodicity are known as general solution
To find a general solution we will use $n$ as an integer where $n\in z$ and $z$ is a real number.
The defining general solutions of the trigonometric functions involve following solutions.

Equation	Solutions
$\sin x=0$	$x=n\pi$
$\cos x=0$	$x=(2n+1)\dfrac{\pi }{2}$
$\tan x=0$	$x=n\pi$
$\sin x=1$	$x=(2n\pi +\pi /2)=(4n+1)\pi /2$
$\cos x=1$	$x=2n\pi$
$\sin x=\sin \alpha$	$x=n\pi +{{(-1)}^{n}}\alpha ,$ where $\alpha \in \left[ \dfrac{-\pi }{2},\dfrac{\pi }{2} \right]$
$\cos x=\cos \alpha$	$x=2n\pi \pm \alpha ,$ where $\alpha \in (0,\,\pi )$
$\tan x=\tan \alpha$	$x=n\pi +\alpha ,$ where $\alpha \in \left( \dfrac{\pi }{2},\,\dfrac{\pi }{2} \right)$
${{\sin }^{2}}x={{\sin }^{2}}\alpha$	$x=n\pi \pm \alpha$
${{\cos }^{2}}x={{\cos }^{2}}\alpha$	$x=n\pi \pm \alpha$
${{\tan }^{2}}x={{\tan }^{2}}\alpha$	$x=n\pi \pm \alpha$

Another method of solving this question is as follows given function in the question is $\sin (x+\pi )=0.5$
Therefore $\sin (x+\pi )=\dfrac{1}{2}$
We kow that $\sin \left( \dfrac{\pi }{6} \right)=\dfrac{1}{2}$
$\sin (x+\pi )=\sin \left( \dfrac{\pi }{6} \right)$
Now, squaring above equation to both side we get
${{\sin }^{2}}(x+\pi )={{\sin }^{2}}\left( \dfrac{\pi }{6} \right)$
The general solution for ${{\sin }^{2}}x={{\sin }^{2}}\alpha$ is $x=n\pi \pm \alpha$
Therefore, we have
$x+\pi =n\pi \pm \alpha$
As $\alpha =\dfrac{\pi }{6}$
Therefore,
$x+\pi =n\pi \pm \dfrac{\pi }{6}$
Now, solve for positive sign
$x+\pi =n\pi +\dfrac{\pi }{6}$
$x=n\pi +\dfrac{\pi }{6}-\pi$
Now, subtract $\pi$ from $\dfrac{\pi }{6}$ we get
$x=n\pi -\dfrac{5\pi }{6}$
Now, solving for negative sign
$x+\pi =n\pi -\dfrac{\pi }{6}$
$x=n\pi -\dfrac{\pi }{6}-\pi$
After solving above equation we get
$x=n\pi -\dfrac{7\pi }{6}$
Therefore, the general solution is
$x=n\pi -\dfrac{5\pi }{6}$ or $x=n\pi -\dfrac{7\pi }{6}$