Question

Find the number of solutions of \[{{\sin }^{2}}x-\sin x-1=0\] in \[\left[ -2\pi ,2\pi \right]\]

Answer 1

Hint: We solve this problem by finding the roots of a given equation using the formula of roots. The formula of roots of quadratic equation \[a{{x}^{2}}+bx+c=0\] is given as
\[x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\]
After finding the value of \[\sin x\] using the above formula we can get the number of solutions using graphs that is the number of solutions of \[f\left( x \right)=g\left( x \right)\] is nothing but number of point of intersections of graph \[y=f\left( x \right)\] and \[y=g\left( x \right)\]

Complete step by step answer:
We are given that the quadratic equation as
\[\Rightarrow {{\sin }^{2}}x-\sin x-1=0\]
Let us assume that \[\sin x=t\] then by substituting it in the above equation we get
\[\Rightarrow {{t}^{2}}-t-1=0\]
We know that the formula of roots of quadratic equation \[a{{x}^{2}}+bx+c=0\] is given as
\[x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\]
By using the above formula to given equation we get
\[\begin{align}
  & \Rightarrow t=\dfrac{-\left( -1 \right)\pm \sqrt{{{1}^{2}}-4\left( 1 \right)\left( -1 \right)}}{2\left( 1 \right)} \\
 & \Rightarrow t=\dfrac{1\pm \sqrt{5}}{2} \\
\end{align}\]
Here, we can see that \['t'\] has two solutions
Let us take one solution as
\[\Rightarrow \sin x=\dfrac{1+\sqrt{5}}{2}\]
Here, we can see that the value that is
\[\Rightarrow \sin x=\dfrac{1+\sqrt{5}}{2}>1\]
Bu we know that the range of \[\sin x\] is \[\left[ -1,1 \right]\]
Therefore, we can say that
\[\Rightarrow \sin x\ne \dfrac{1+\sqrt{5}}{2}\]
Now let us take the second solution
\[\Rightarrow \sin x=\dfrac{1-\sqrt{5}}{2}\]
We know that the number of solutions of \[f\left( x \right)=g\left( x \right)\] is nothing but number of point of intersections of graph \[y=f\left( x \right)\] and \[y=g\left( x \right)\]
Now let us draw the graph of \[y=\sin x\] and \[y=\dfrac{1-\sqrt{5}}{2}\] in the domain \[\left[ -2\pi ,2\pi \right]\] as

Here, we can see that red line is \[y=\sin x\] and green line is \[y=\dfrac{1-\sqrt{5}}{2}\]
We can see that both graphs intersect at 4 points in the domain \[\left[ -2\pi ,2\pi \right]\]

Therefore, the number of solutions of \[{{\sin }^{2}}x-\sin x-1=0\] is 4.

Note: We can solve this problem in another method.
We are given the equation as
\[\begin{align}
  & \Rightarrow {{\sin }^{2}}x-\sin x-1=0 \\
 & \Rightarrow \sin x=-\left( 1-{{\sin }^{2}}x \right).........equation(i) \\
\end{align}\]
We know that the standard result that is
\[\begin{align}
  & \Rightarrow {{\sin }^{2}}x+{{\cos }^{2}}x=1 \\
 & \Rightarrow {{\cos }^{2}}x=1-{{\sin }^{2}}x \\
\end{align}\]
By substituting this value in equation (i) we get
\[\Rightarrow \sin x=-{{\cos }^{2}}x\]
Now let us draw the graph of \[y=\sin x\] and \[y=-{{\cos }^{2}}x\] in the domain \[\left[ -2\pi ,2\pi \right]\] as

Here, we can see that red line is \[y=\sin x\] and blue line is \[y=-{{\cos }^{2}}x\]
We can see that both graphs intersect at 4 points in the domain \[\left[ -2\pi ,2\pi \right]\]
Therefore, the number of solutions of \[{{\sin }^{2}}x-\sin x-1=0\] is 4.