Question

How do you solve the equation \[4{{\sin }^{2}}\theta -4\sin \theta +1=0\]?

Answer 1

Hint: This type of problem is based on the concept of finding the value of \[\theta \] from the trigonometric table. First, we have to substitute \[\sin \theta =u\] and obtain an equation with variable u. Then, consider the obtained equation and use quadratic formula to find the value of ‘u’, that is, \[u=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\]. Here, a=4, b=-4 and c=1. We need to find the values of the given equation by making necessary calculations. And then after finding the value of u, substitute \[u=\sin \theta \] and take \[{{\sin }^{-1}}\] on both the sides of the expression. Then, we need to find the value of \[\theta \] which is the required answer.

Complete step-by-step solution:
According to the question, we are asked to solve the given equation \[4{{\sin }^{2}}\theta -4\sin \theta +1=0\].
We have been given the equation is \[4{{\sin }^{2}}\theta -4\sin \theta +1=0\]. -----(1)
Let us substitute \[\sin \theta \] as u, that is \[\sin \theta =u\].
Therefore, equation (1) becomes
\[4{{u}^{2}}-4u+1=0\] -----------(2)
Now, we have obtained a quadratic equation with a variable.
We know that for a quadratic equation \[a{{x}^{2}}+bx+c=0\],
\[x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\].
Here, by comparing this with equation (2), we get,
a=4, b=-4 and c=1.
 And also \[u=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\].
Therefore,
\[u=\dfrac{-\left( -4 \right)\pm \sqrt{{{\left( -4 \right)}^{2}}-4\left( 4 \right)\left( 1 \right)}}{2\left( 4 \right)}\]
\[\Rightarrow u=\dfrac{4\pm \sqrt{{{4}^{2}}-4\left( 4 \right)}}{8}\]
We know that the square of 4 is 16.
Using this in the above obtained expression, we get,
\[u=\dfrac{4\pm \sqrt{16-4\times 4}}{8}\]
On further simplifications, we get,
\[\Rightarrow u=\dfrac{4\pm \sqrt{16-16}}{8}\]
\[\Rightarrow u=\dfrac{4\pm \sqrt{0}}{8}\]
We know that \[\sqrt{0}=0\]. Therefore, we get
\[u=\dfrac{4}{8}\]
\[\Rightarrow u=\dfrac{4}{4\times 2}\]
Since 4 is common in both numerator and denominator, let us cancel the common term 4.
\[\Rightarrow u=\dfrac{1}{2}\]
But we have assumed \[u=\sin \theta \].
\[\therefore \sin \theta =\dfrac{1}{2}\]
Let us take \[{{\sin }^{-1}}\] on both the sides of the expression, we get
\[{{\sin }^{-1}}\left( \sin \theta \right)={{\sin }^{-1}}\left( \dfrac{1}{2} \right)\]
Using the trigonometric property for inverse \[{{\sin }^{-1}}\left( \sin \theta \right)=\theta \], we get
\[\Rightarrow \theta ={{\sin }^{-1}}\left( \dfrac{1}{2} \right)\]
We know that \[\sin \left( \dfrac{\pi }{6} \right)=\dfrac{1}{2}\].
But from the trigonometric cycle, we understand that after an interval of \[2\pi \], the same value for sin function repeats.
Therefore, for \[\dfrac{\pi }{6}+2n\pi \], the value of sin function is \[\dfrac{1}{2}\].
Hence, the values of \[\theta \] for the equation \[4{{\sin }^{2}}\theta -4\sin \theta +1=0\] are \[\dfrac{\pi }{6}+2n\pi \] where n is an integer.

Note: Whenever you get this type of problem, we should always try to make the necessary changes in the given equation by substitution to get a quadratic equation. We should avoid calculation mistakes based on sign conventions. We should always make some necessary calculations to obtain zero in the right-hand side of the equation for easy calculations. Also we can solve this problem by not substituting \[\sin \theta =u\] and considering \[\sin \theta \] being the variable in the quadratic equation.