How do you solve $$\sin x = \dfrac{{ - \sqrt 2 }}{2}$$ ?

Question

Vedantu Content Team · Accepted Answer

Hint: In this question we have to find the value $$x$$ for the given trigonometric function, first factorise the denominator and then using the trigonometric ratio table and using the fact that sin positive in first and second quadrant and negative in third and fourth quadrant and negative in third and fourth quadrant, we will get the required answer.Complete step by step solution:Given expression is $$\sin x = \dfrac{{ - \sqrt 2 }}{2}$$ ,Now we know that $$2 = \sqrt 2  \times \sqrt 2 $$ , now substituting the value in the given expression we get,$$ \Rightarrow \sin x = \dfrac{{ - \sqrt 2 }}{{\sqrt 2  \times \sqrt 2 }}$$ ,Now simplifying we get,$$ \Rightarrow \sin x = \dfrac{{ - 1}}{{\sqrt 2 }}$$ ,Now taking sin inverse on both sides we get,$$ \Rightarrow {\sin ^{ - 1}}\left( {\sin x} \right) = {\sin ^{ - 1}}\left( {\dfrac{{ - 1}}{{\sqrt 2 }}} \right)$$ ,Now we know that $$\sin \left( {{{\sin }^{ - 1}}x} \right) = x$$ , now applying the fact we get,$$ \Rightarrow x = {\sin ^{ - 1}}\left( {\dfrac{{ - 1}}{{\sqrt 2 }}} \right)$$ ,TRIGONOMETRIC RATIO $${0^O}$$  $${30^o}$$  $${45^o}$$  $${60^o}$$  $${90^o}$$  $${120^o}$$  $${135^o}$$  $${180^o}$$  $${225^o}$$  $${270^o}$$  $${315^o}$$  $${360^o}$$ 0 $$\dfrac{\pi }{6}$$  $$\dfrac{\pi }{4}$$  $$\dfrac{\pi }{3}$$  $$\dfrac{\pi }{2}$$  $$\dfrac{{2\pi }}{3}$$  $$\dfrac{{3\pi }}{4}$$  $$\pi $$  $$\dfrac{{5\pi }}{4}$$  $$\dfrac{{3\pi }}{2}$$  $$\dfrac{{7\pi }}{4}$$  $$2\pi $$ Sin0 $$\dfrac{1}{2}$$  $$\dfrac{1}{{\sqrt 2 }}$$  $$\dfrac{{\sqrt 3 }}{2}$$ 1 $$\dfrac{{\sqrt 3 }}{2}$$  $$\dfrac{1}{{\sqrt 2 }}$$ 0 $$\dfrac{{ - 1}}{{\sqrt 2 }}$$ -1 $$\dfrac{{ - 1}}{{\sqrt 2 }}$$ 0Cos1 $$\dfrac{{\sqrt 3 }}{2}$$  $$\dfrac{1}{{\sqrt 2 }}$$  $$\dfrac{1}{2}$$ 0 $$\dfrac{{ - 1}}{2}$$  $$\dfrac{{ - 1}}{{\sqrt 2 }}$$ -1 $$\dfrac{{ - 1}}{{\sqrt 2 }}$$ 0 $$\dfrac{1}{{\sqrt 2 }}$$ 1Tan0 $$\dfrac{1}{{\sqrt 3 }}$$ 1 $$\sqrt 3 $$ undefined $$\dfrac{{ - 1}}{{\sqrt 3 }}$$ -101undefined-10From the trigonometric table we can say that,$$ \Rightarrow x = \dfrac{{5\pi }}{4},\dfrac{{7\pi }}{4}$$ ,We also know that fact that sin positive in first and second quadrant and negative in third and fourth quadrant and negative in third and fourth quadrant,So we can rewrite as,$$ \Rightarrow $$  $$x = \left( {\pi  - \left( { - \dfrac{\pi }{4}} \right)} \right)$$ and $$x = \left( {2\pi  - \left( {\dfrac{\pi }{4}} \right)} \right)$$ ,Now simplifying we get,$$ \Rightarrow x = \dfrac{{5\pi }}{4}$$ and $$x = \left( {\dfrac{{7\pi }}{4}} \right)$$ ,Now generalising the answer we get,$$ \Rightarrow $$  $ x = 2\pi n + \dfrac{{5\pi }}{4} $ and $ x = 2\pi n + \dfrac{{7\pi }}{4} $ ,So, by solving the given function we get the values of $ x $ as $ 2\pi n + \dfrac{{5\pi }}{4} $ and  $ 2\pi n + \dfrac{{7\pi }}{4} $ .The value of  $ x $ when the given function $ \sin x = \dfrac{{ - \sqrt 2 }}{2} $ is solved will be equal to $ 2\pi n + \dfrac{{5\pi }}{4} $ and  $ 2\pi n + \dfrac{{7\pi }}{4} $ .Note: The trigonometry and inverse trigonometry are inverse of each other, and they are represented by arc of the function or the function raised to the power of -1. The sine, cosine, tangent, cosecant, secant, and cotangent are the six types of ratios of trigonometry. The inverse trigonometric functions in trigonometry are used to calculate the angles through any of the trigonometric ratios.

How do you solve \[\sin x = \dfrac{{ - \sqrt 2 }}{2}\] ?