Question

How do you solve $2\sin x+1=0?$

Answer 1

Hint: The given function in the question is the trigonometric function of the sine function. The domain and the range of the sine function is all real numbers and the range is $-1\le \theta \le 1$.
The inverse of the sine function is the cosec function $\sin x=\dfrac{1}{\csc x}$.
In the first quadrant, the sine function increase $0$to $1$
In the second quadrant, the sine function decreases by $1$ to $0$
In the third quadrant, the sine function decreases by $0$ to $-1$
In the fourth quadrant, the sine function increase by $-1$ to $0$
The values of some $\theta $ will help to simplify the sine function-

$\theta $	${{0}^{0}}$	${{30}^{0}}$	\[{{45}^{0}}\]	${{60}^{0}}$	${{90}^{0}}$	${{120}^{0}}$	${{135}^{0}}$	${{150}^{0}}$	${{180}^{0}}$
Sine	$0$	${}^{1}/{}_{2}$	${}^{1}/{}_{\sqrt{2}}$	${\sqrt{3}}/{2}\;$	$1$	${\sqrt{3}}/{2}\;$	${}^{1}/{}_{\sqrt{2}}$	${}^{1}/{}_{2}$	$0$

Complete step by step solution:
The given function is $2\sin x+1=0$
To simplify it we will add $-1$ in the left side and the right side of the function then it will be written as
$\Rightarrow 2\sin x+1+(-1)=-1$
$\Rightarrow 2\sin x=-1$
To more simplify the above term we will divide in the left side and right side of the function by $2$
\[\Rightarrow \dfrac{2\sin x}{2}=\dfrac{-1}{2}\]
The above term will be \[\sin x=\dfrac{-1}{2}\]
$\Rightarrow x={{\sin }^{-1}}\left( \dfrac{-1}{2} \right)$
$\Rightarrow x=-{{\sin }^{-1}}\left( \dfrac{1}{2} \right)$
$\Rightarrow x=-\dfrac{\pi }{6}$
As we know that in the third and fourth quadrant it will be negative.
In the third quadrant is will be $\pi +\dfrac{\pi }{6}=\dfrac{7\pi }{6}$
And in the fourth quadrant, it will be $2\pi -\dfrac{\pi }{6}=\dfrac{11\pi }{6}$

Hence, the answer will be $-\dfrac{\pi }{6}$ or $\dfrac{7\pi }{6}$and $\dfrac{11\pi }{6}$

Note:
According to the ‘ASTC’ rule, the sine function always gets a positive value in the first quadrant and the second quadrant.
The sine function is commonly used in periodic motion, light waves, sound, etc.