
How do you solve \[\dfrac{1}{\sin x}\]= $ 2 $ ?
Answer
547.2k+ views
Hint: This type of question involves the concept of trigonometric ratios and angles. In this question, we need to find out the value of $ \sin x $. For this, we will firstly cross-multiply the LHS and RHS and then find the value of $ \sin x $ by undergoing basic mathematical calculations.
Complete step by step answer:
The given equation is \[\dfrac{1}{\sin x}\]= $ 2 $ .
In the above equation R.H.S $ 2 $ can be written as $ \dfrac{2}{1} $ .
$ \Rightarrow $ $ \dfrac{1}{\sin x}=\dfrac{2}{1} $
Now, on cross multiplying the L.H.S and R.H.S, we get
$ \begin{align}
& \Rightarrow \,2\left(\sin x \right)\,=\,1 \\
& \Rightarrow \,\sin x\,=\,\dfrac{1}{2}\,\,\,\left( i \right) \\
\end{align} $
Also, using the basic trigonometric angles, we know that $ \sin {{30}^{\circ }}=\dfrac{1}{2} $ and also $ \sin {{150}^{\circ }}=\dfrac{1}{2} $ .
Substituting the value of $ \dfrac{1}{2} $ as $ \sin {{30}^{\circ }} $ in the R.H.S of equation $ (i) $ we get
$ \begin{align}
& \sin x=\sin {{30}^{\circ }} \\
& \therefore x={{30}^{\circ }} \\
\end{align} $
Now on substituting the value of $ \dfrac{1}{2} $ as $ \sin {{150}^{\circ }} $ in the R.H.S of equation $ (i) $ we get
$ \begin{align}
& \sin x=\sin {{150}^{\circ }} \\
& \therefore x={{150}^{\circ }} \\
\end{align} $
Hence in this question there can be two values of $ x $ as $ x={{30}^{\circ }} $ and $ x={{150}^{\circ }} $ .
Therefore the $ x $ can have both the values and this is a common scenario in the case of the trigonometric ratios and angles questions as in any given situation two different trigonometric angles can have the same values and hence the values of variables can be two different angles.
Note:
Always remain careful while solving the trigonometric angles and if the solution has two different values of the angles then mention both the values. Often students commit mistakes in solving the question and end up mentioning only one value. This creates a problem in the MCQ type question when only one of the values will be given as the correct answer. Hence if they haven’t found both the values they may get confused in the question. So Always see whether a given trigonometric angle has one or more than one value.
Complete step by step answer:
The given equation is \[\dfrac{1}{\sin x}\]= $ 2 $ .
In the above equation R.H.S $ 2 $ can be written as $ \dfrac{2}{1} $ .
$ \Rightarrow $ $ \dfrac{1}{\sin x}=\dfrac{2}{1} $
Now, on cross multiplying the L.H.S and R.H.S, we get
$ \begin{align}
& \Rightarrow \,2\left(\sin x \right)\,=\,1 \\
& \Rightarrow \,\sin x\,=\,\dfrac{1}{2}\,\,\,\left( i \right) \\
\end{align} $
Also, using the basic trigonometric angles, we know that $ \sin {{30}^{\circ }}=\dfrac{1}{2} $ and also $ \sin {{150}^{\circ }}=\dfrac{1}{2} $ .
Substituting the value of $ \dfrac{1}{2} $ as $ \sin {{30}^{\circ }} $ in the R.H.S of equation $ (i) $ we get
$ \begin{align}
& \sin x=\sin {{30}^{\circ }} \\
& \therefore x={{30}^{\circ }} \\
\end{align} $
Now on substituting the value of $ \dfrac{1}{2} $ as $ \sin {{150}^{\circ }} $ in the R.H.S of equation $ (i) $ we get
$ \begin{align}
& \sin x=\sin {{150}^{\circ }} \\
& \therefore x={{150}^{\circ }} \\
\end{align} $
Hence in this question there can be two values of $ x $ as $ x={{30}^{\circ }} $ and $ x={{150}^{\circ }} $ .
Therefore the $ x $ can have both the values and this is a common scenario in the case of the trigonometric ratios and angles questions as in any given situation two different trigonometric angles can have the same values and hence the values of variables can be two different angles.
Note:
Always remain careful while solving the trigonometric angles and if the solution has two different values of the angles then mention both the values. Often students commit mistakes in solving the question and end up mentioning only one value. This creates a problem in the MCQ type question when only one of the values will be given as the correct answer. Hence if they haven’t found both the values they may get confused in the question. So Always see whether a given trigonometric angle has one or more than one value.
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