Question

When \[\sin x = 1\] what does \[x\] equal ?

Answer 1

Hint: In this given question, we need to find the value of \[x\]. Given that \[\sin x=1\] . The basic trigonometric functions are sine , cosine and tangent. Cosine is nothing but a ratio of the adjacent side of a right angle to the hypotenuse of the right angle . Here we need to find the value of \[x\] when \[\sin x=1\] . With the help of the Trigonometric functions , we can find the value of \[x\].
Complete step by step solution :

Complete step by step solution:
Given, \[\sin x = 1\]
We need to find the value of \[x\].
By taking inverse of sine on both sides of the given expression,
We get,
\[sin^{- 1}(\sin x)\ = \sin^{- 1}(1)\]
On simplifying,
We get,
\[x = sin^{- 1}(1)\]
\[\Rightarrow x=\dfrac{\pi}{2}\]
If we restrict our answer \[x\] within \[[0,2π]\]
The period of sine function is \[2π\] . So values will report every \[2π\] radians in both directions.
\[\Rightarrow \sin(\dfrac{\pi}{2}+(2n\pi))\ = 1\] for any integer \[n\] including \[0\] . \[(n\ \in \ Z)\]
That is \[x =\dfrac{\pi}{2}+2n\pi\] where \[n \in Z\]
By substituting \[n = 0\] ,
We get, \[x = \dfrac{\pi}{2}\]
Whereas \[\sin\dfrac{\pi}{2}=1\]
Hence the value of \[x = 90^{o},\ 450^{o}\], and so on.
Therefore \[x = \dfrac{\pi}{2}+2n\pi\] where \[n \in Z\]
Final answer :
When \[\sin x = 1\] , the value of \[x =\dfrac{\pi}{2}+2n\pi\] where \[n \in Z\]

Note: The concept used in this problem is trigonometric identities and ratios. Trigonometric identities are nothing but they involve trigonometric functions including variables and constants. The common technique used in this problem is the use of trigonometric functions . Trigonometric functions are also known as circular functions or geometrical functions. The inverse of the sine function is nothing but it is a sine function raised to the power of \[(-1)\] . Sine functions are positive in the first and second quadrant.