Question

Find the general solution of $\csc \theta =-\sqrt{2.}$

Answer 1

Hint: Get the given relation in the problem in terms of $\sin \theta $, using the relation $\sin \theta =\dfrac{1}{\csc \theta }$. Now, get the evaluated equation in form of $\sin x=\sin y$ and hence, use the general solution of $\sin x=\sin y$ to get the general solution of the formed equation as: $x=n\pi +\left( -1 \right)y$.

Complete step-by-step answer:

Use the following result to solve the problem:

$\begin{align}

  & \to \sin \left( \pi -A \right)=-\operatorname{sinA} \\

 & \to \sin \dfrac{\pi }{4}=\dfrac{1}{\sqrt{2}} \\

\end{align}$


We have

$\csc \theta =-\sqrt{2}$ ………………(i)

As, we know the relation between $\sin \theta ,\csc \theta $ is given as

$\sin \theta =\dfrac{1}{\csc \theta }$ …………….(ii)

So, we can get a familiar relation i.e. in terms of $\sin \theta $, So, we can use the equation

(ii) to get the given expression in the problem in terms of $\sin \theta $ as

$\sin \theta =\dfrac{1}{-\sqrt{2}}=\dfrac{-1}{\sqrt{2}}$

$\sin \theta =\dfrac{-1}{\sqrt{2}}$………….(iii)

Now, as we know sine function is positive in the first and second quadrant and negative in the third and fourth quadrant. So, we can write the trigonometric relation for sine function with respect to third quadrant as

$\sin \left( \pi +A \right)=-\sin A$ ……………..(iv)

And we know very well that sine function will give value of $\dfrac{\text{1}}{\sqrt{\text{2}}}\to \text{ angle }\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}}$. So, on putting $A=\dfrac{\pi }{4}$ to the equation (iv), we get

$\sin \left( \pi +\dfrac{\pi }{4} \right)=-\sin \dfrac{\pi }{4}$

Hence, put $\sin \dfrac{\pi }{4}=\dfrac{1}{\sqrt{2}}$, we get

$\sin \dfrac{5\pi }{4}=-\dfrac{1}{\sqrt{2}}$ ………………….(v)

Now, we can put $\dfrac{-1}{\sqrt{2}}$ in the equation (iii) as $\sin \dfrac{5\pi }{4}$ using the equation (v).

So, we can rewrite the equation (iii).

Now, as we know the general solution of equation $\sin x=\sin y$ can be given as

If $\sin x=\sin y$, then

$x=n\pi +{{\left( -1 \right)}^{n}}y$ …………………(vii)

Where, $n\in z\to n$ is an integer.

Now, we can compare the equation (vi) with the equation $\sin x=\sin y$ and hence, we can get values of x, y, so that the general solution can be calculated with the equation (vi). So, on comparing $\sin \theta =\sin \left( \dfrac{5\pi }{4} \right)$ with the equation $\sin x=\sin y$, we get

$x=\theta ,y=\dfrac{5\pi }{4}$

Now, we can substitute these values to the equation (vi) to get the general solution of equation (i). so, we get

$\theta =n\pi +{{\left( -1 \right)}^{n}}\dfrac{5\pi }{4}$

 Where $n\in z$.

Now, we can put number of values of n to get solutions as

$\sin \theta =sin\dfrac{5\pi }{4}$ …………………(vi)


Note: One may prove the general solution of $\sin x=\sin y$ by following ways:

$\sin x-\sin y=0$


Apply $\operatorname{sinC}-sinD=2sin\dfrac{C-D}{2}\cos \dfrac{C+D}{2}$


So, we get


$2\sin \dfrac{x-y}{2}\cos \dfrac{x+y}{2}=0$


Now, put $\sin \left( \dfrac{x-y}{2} \right)=0,\cos \left( \dfrac{x+y}{2} \right)=0$


And apply the general solution of equations $\sin \theta =0,\cos \theta =0\to \theta =n\pi


,\theta =\left( 2n+1 \right)\dfrac{\pi }{2}$ respectively. Hence, get the general solution for


$\sin x=\sin y$ as


$\sin \theta =\sin \dfrac{5\pi }{4}$ …………….(vi)


Now, as we know the


$\theta =n\pi +{{\left( -1 \right)}^{n}}\dfrac{5\pi }{4}$


Another approach to write $\sin \theta =-\dfrac{1}{\sqrt{2}}$ can be given as


$\sin \theta =\sin \left( -\dfrac{\pi }{4} \right)$


Where, we know $\sin \left( -x \right)=-\sin $


Now, use $x=n\pi +{{\left( -1 \right)}^{n}}y$ to get a general solution. So, we get


$\theta =n\pi +{{\left( -1 \right)}^{n}}\left( -\dfrac{\pi }{4} \right)$

Now, one may get confused with the two different general solutions. So, we need to know that there are infinite ways of representing the general solution of any trigonometric relation, but solutions will be the same, on putting values of n = 0, 1, 2, 3……………… hence, only the representation will differ by another approach, not solutions.