How do you solve \[\sin \,x + 2 = 3\]

Hint: Here the question is related to the trigonometry, we use the trigonometry ratios and we are to solve this question. In this question we can find the value of x such that it satisfies the given question. The value of x is determined with the help of angle of trigonometry ratios table.

Complete step by step solution:
The question is related to trigonometry and it includes the trigonometry ratios. The trigonometry ratios are sine, cosine and tan.
Now consider the given trigonometric equation
 \[\sin \,x + 2 = 3\]
Now take 2 from LHS to RHS and we get
 \[ \Rightarrow \sin \,x = 3 - 2\]
On simplifying we have
 \[ \Rightarrow \sin \,x = 1\]
To determine the value of x we use the inverse of the trigonometric function.
 \[ \Rightarrow x = {\sin ^{ - 1}}(1)\]
By the table for the trigonometry ratios for standard angles it is given as
 \[ \Rightarrow x = \dfrac{\pi }{2}\]
The exact value of \[{\sin ^{ - 1}}(1)\] is \[\dfrac{\pi }{2}\]
Hence we have solved and simplified the given trigonometric function.
As we know about the ASTC rule the sin is negative in the third quadrant and the fourth quadrant. To find the solution add the reference angle and \[2\pi \]
 \[
   \Rightarrow x = \dfrac{\pi }{2} + 2\pi \\
   \Rightarrow x = \dfrac{{5\pi }}{2} \;
 \]
The period of \[\sin x\] functions is \[\pi \] so values will repeat every \[\pi \] radians in both directions.
 \[x = \dfrac{\pi }{2} + 2\pi n\] , for any integer n.
So, the correct answer is “ \[x = \dfrac{\pi }{2} + 2\pi n\] ”.

Note: Since we can say at which value of x the addition of cosine and sine will be zero. The ASTC rule defined as all sine tan cosine this explains all trigonometry ratios are positive in the first quadrant. Sine trigonometry ratio is positive in the second quadrant. Tan trigonometry ratio is positive in the third quadrant. The cosine trigonometry ratio is positive in the fourth quadrant.