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Question

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

When sinx=1\sin x = 1 what does xx equal ?

Explanation

Solution

In this given question, we need to find the value of xx. Given that sinx=1\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 xx when sinx=1\sin x=1 . With the help of the Trigonometric functions , we can find the value of xx.
Complete step by step solution :

Complete step by step solution:
Given, sinx=1\sin x = 1
We need to find the value of xx.
By taking inverse of sine on both sides of the given expression,
We get,
sin1(sinx) =sin1(1)sin^{- 1}(\sin x)\ = \sin^{- 1}(1)
On simplifying,
We get,
x=sin1(1)x = sin^{- 1}(1)
x=π2\Rightarrow x=\dfrac{\pi}{2}
If we restrict our answer xx within [0,2π][0,2π]
The period of sine function is 2π . So values will report every 2π radians in both directions.
sin(π2+(2nπ)) =1\Rightarrow \sin(\dfrac{\pi}{2}+(2n\pi))\ = 1 for any integer nn including 00 . (n  Z)(n\ \in \ Z)
That is x=π2+2nπx =\dfrac{\pi}{2}+2n\pi where nZn \in Z
By substituting n=0n = 0 ,
We get, x=π2x = \dfrac{\pi}{2}
Whereas sinπ2=1\sin\dfrac{\pi}{2}=1
Hence the value of x=90o, 450ox = 90^{o},\ 450^{o}, and so on.
Therefore x=π2+2nπx = \dfrac{\pi}{2}+2n\pi where nZn \in Z
Final answer :
When sinx=1\sin x = 1 , the value of x=π2+2nπx =\dfrac{\pi}{2}+2n\pi where nZn \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)(-1) . Sine functions are positive in the first and second quadrant.