Question

If $$\sin^{2} x-2\sin x-1=0$$ has exactly four different solutions in $$x\in \left[ 0,n\pi \right] $$, then minimum value of n can be $$\left( n\in \mathrm{N} \right) $$.

Answer 1

Hint: In this question it is given that if $$\sin^{2} x-2\sin x-1=0$$ has exactly four different solutions in $$x\in \left[ 0,n\pi \right] $$, we have to find the minimum value of n, where $$\left( n\in \mathrm{N} \right) $$. So to find the solution we have to know one formula, which says that,
if $\sin x=\sin y$ then $$x=n\pi +\left( -1\right)^{n} y$$......(1)
Where $$n=0,\pm 1,\pm 2,\cdots$$.
Complete step-by-step solution:
Given equation,
$$\sin^{2} x-2\sin x-1=0$$
$$\Rightarrow \sin^{2} x-2\sin x=1$$
$$\Rightarrow \sin^{2} x-2\sin x+1=1+1$$ [adding 1 on the both side]
$$\Rightarrow \sin^{2} x-2\sin x+1=2$$
$$\Rightarrow \left( \sin x\right)^{2} -2\cdot \sin x\cdot 1+1^{2}=2$$
Now as we know that $$a^{2}+2ab+b^{2}=\left( a+b\right)^{2} $$, so by using this identity we can write the above equation as,
$$\Rightarrow \left( \sin x-1\right)^{2} =2$$
$$\Rightarrow \sin x-1=\pm \sqrt{2}$$
$$\Rightarrow \sin x=1\pm \sqrt{2}$$
Either,
$$\Rightarrow \sin x=1+\sqrt{2}$$ ……..(2)
Or,
$$\Rightarrow \sin x=1-\sqrt{2}$$...........(3)

Since, as we know that $$-1\leq \sin x\leq 1$$, so equation(2) is not possible.
$$\therefore \sin x=1-\sqrt{2}$$
$$\therefore \sin x=-\left( \sqrt{2} -1\right) $$........(4)
Now let $$\sin y=\left( \sqrt{2} -1\right) $$
Therefore, from equation (4) we get,
$$\sin x=-\sin y$$
$$\sin x=\sin \left( -y\right) $$[$$\because -\sin \theta =\sin \left( -\theta \right) $$]
$$\Rightarrow x=n\pi +\left( -1\right)^{n} \left( -y\right) $$ [using equation (1)]
Where $$n=0,\pm 1,\pm 2,\cdots$$.
Now, by putting the values of ‘n’ we get,
$$\Rightarrow x=\pi +y,\ 2\pi -y,\ 3\pi +y,\ 4\pi -y,\ 5\pi +y$$
Since, $$x\in \left[ 0,n\pi \right] $$
So the possible values of x are $$\pi +y,\ 2\pi -y,\ 3\pi +y,\ 4\pi -y,$$
Therefore, we can say that for exactly four different solutions the minimum value of n is 4.
So the answer is 4.
Note: While finding the possible values of x, we have to put n=0, then n=1, then -1 and so on, but we only take those values which lies in $$\left[ 0,2\pi \right] $$, and you have to keep in mind that every values of x which is lies in $$\left[ 0,2\pi \right] $$ is positive, so because of that we have to exclude those values which is generated by n=0,-1,-2,-3,....