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Question: If \[\sin^{2} x-2\sin x-1=0\] has exactly four different solutions in \[x\in \left[ 0,n\pi \right] \...

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

Explanation

Solution

Hint: In this question it is given that if sin⁡2x−2sin⁡x−1=0\sin^{2} x-2\sin x-1=0 has exactly four different solutions in x∈[0,nπ]x\in \left[ 0,n\pi \right] , we have to find the minimum value of n, where (n∈N)\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\sin x=\sin y then x=nπ+(−1)nyx=n\pi +\left( -1\right)^{n} y......(1)
Where n=0,±1,±2,⋯n=0,\pm 1,\pm 2,\cdots.
Complete step-by-step solution:
Given equation,
sin⁡2x−2sin⁡x−1=0\sin^{2} x-2\sin x-1=0
⇒sin⁡2x−2sin⁡x=1\Rightarrow \sin^{2} x-2\sin x=1
⇒sin⁡2x−2sin⁡x+1=1+1\Rightarrow \sin^{2} x-2\sin x+1=1+1 [adding 1 on the both side]
⇒sin⁡2x−2sin⁡x+1=2\Rightarrow \sin^{2} x-2\sin x+1=2
⇒(sin⁡x)2−2⋅sin⁡x⋅1+12=2\Rightarrow \left( \sin x\right)^{2} -2\cdot \sin x\cdot 1+1^{2}=2
Now as we know that a2+2ab+b2=(a+b)2a^{2}+2ab+b^{2}=\left( a+b\right)^{2} , so by using this identity we can write the above equation as,
⇒(sin⁡x−1)2=2\Rightarrow \left( \sin x-1\right)^{2} =2
⇒sin⁡x−1=±2\Rightarrow \sin x-1=\pm \sqrt{2}
⇒sin⁡x=1±2\Rightarrow \sin x=1\pm \sqrt{2}
Either,
⇒sin⁡x=1+2\Rightarrow \sin x=1+\sqrt{2} ……..(2)
Or,
⇒sin⁡x=1−2\Rightarrow \sin x=1-\sqrt{2}...........(3)

Since, as we know that −1≤sin⁡x≤1-1\leq \sin x\leq 1, so equation(2) is not possible.
∴sin⁡x=1−2\therefore \sin x=1-\sqrt{2}
∴sin⁡x=−(2−1)\therefore \sin x=-\left( \sqrt{2} -1\right) ........(4)
Now let sin⁡y=(2−1)\sin y=\left( \sqrt{2} -1\right)
Therefore, from equation (4) we get,
sin⁡x=−sin⁡y\sin x=-\sin y
sin⁡x=sin⁡(−y)\sin x=\sin \left( -y\right) [∵−sin⁡θ=sin⁡(−θ)\because -\sin \theta =\sin \left( -\theta \right) ]
⇒x=nπ+(−1)n(−y)\Rightarrow x=n\pi +\left( -1\right)^{n} \left( -y\right) [using equation (1)]
Where n=0,±1,±2,⋯n=0,\pm 1,\pm 2,\cdots.
Now, by putting the values of ‘n’ we get,
⇒x=π+y, 2π−y, 3π+y, 4π−y, 5π+y\Rightarrow x=\pi +y,\ 2\pi -y,\ 3\pi +y,\ 4\pi -y,\ 5\pi +y
Since, x∈[0,nπ]x\in \left[ 0,n\pi \right]
So the possible values of x are π+y, 2π−y, 3π+y, 4π−y,\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 [0,2π]\left[ 0,2\pi \right] , and you have to keep in mind that every values of x which is lies in [0,2π]\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,....