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Observe the following statements.Assertion: A: The general solution of $\sin x=-1$ is $n\pi +{{\left( -1 \right)}^{n}}\dfrac{3\pi }{2}$.Reason: R: The principal value of $\sin x=k$ lies in $\left[ -\dfrac{\pi }{2},\dfrac{\pi }{2} \right]$.A. Both A and R are true and R is the correct explanation of A.B. Both A and R are true and R is not the correct explanation of A.C. A is true but R is false.D. A is false but R is true.

Last updated date: 19th Jul 2024
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Hint: For solving this question you should know about finding the values of any general trigonometric function. In this problem we will find the general solution for this as a term of $\alpha$ and for $\sin x$ we will take that as a negative value. And then we get the exact general solution for this and by this we will calculate that the values of this will lie from where to where.

According to our question it is asked of us to observe the given statements and then select the right answer from the options. So, if we see our question, then a general solution of $\sin x=-1$ is given and that is provided as a term. Here we will find our own solution for $\sin x=-1$. And we will get a new solution for $\sin x=-1$. Now, we will compare both the solutions, if both are same, then the assertion A is true, otherwise the assertion is wrong. And we find the interval for which the principle values of $\sin x=k$lies. If the interval is also the same for the desired value, then reason R is also right, otherwise the reason is also wrong.
So, if we find the general solution of $\sin x$, then,
$\sin x=\sin \alpha$ is $x=n\pi +{{\left( -1 \right)}^{n}}\alpha$
$\sin x=-1=\sin \left( -\dfrac{\pi }{2} \right)$
$x=n\pi +{{\left( -1 \right)}^{n}}\left( -\dfrac{\pi }{2} \right)$
Moreover the principal value of $\sin x$ lies in $\left[ -\dfrac{\pi }{2},\dfrac{\pi }{2} \right]$.