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

How do you solve the equation 4sin⁡2θ−4sin⁡θ+1=04{{\sin }^{2}}\theta -4\sin \theta +1=0?

Explanation

Solution

This type of problem is based on the concept of finding the value of θ\theta from the trigonometric table. First, we have to substitute sin⁡θ=u\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=−b±b2−4ac2au=\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⁡θu=\sin \theta and take sin⁡−1{{\sin }^{-1}} on both the sides of the expression. Then, we need to find the value of θ\theta which is the required answer.

Complete step-by-step solution:
According to the question, we are asked to solve the given equation 4sin⁡2θ−4sin⁡θ+1=04{{\sin }^{2}}\theta -4\sin \theta +1=0.
We have been given the equation is 4sin⁡2θ−4sin⁡θ+1=04{{\sin }^{2}}\theta -4\sin \theta +1=0. -----(1)
Let us substitute sin⁡θ\sin \theta as u, that is sin⁡θ=u\sin \theta =u.
Therefore, equation (1) becomes
4u2−4u+1=04{{u}^{2}}-4u+1=0 -----------(2)
Now, we have obtained a quadratic equation with a variable.
We know that for a quadratic equation ax2+bx+c=0a{{x}^{2}}+bx+c=0,
x=−b±b2−4ac2ax=\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=−b±b2−4ac2au=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}.
Therefore,
u=−(−4)±(−4)2−4(4)(1)2(4)u=\dfrac{-\left( -4 \right)\pm \sqrt{{{\left( -4 \right)}^{2}}-4\left( 4 \right)\left( 1 \right)}}{2\left( 4 \right)}
⇒u=4±42−4(4)8\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=4±16−4×48u=\dfrac{4\pm \sqrt{16-4\times 4}}{8}
On further simplifications, we get,
⇒u=4±16−168\Rightarrow u=\dfrac{4\pm \sqrt{16-16}}{8}
⇒u=4±08\Rightarrow u=\dfrac{4\pm \sqrt{0}}{8}
We know that 0=0\sqrt{0}=0. Therefore, we get
u=48u=\dfrac{4}{8}
⇒u=44×2\Rightarrow u=\dfrac{4}{4\times 2}
Since 4 is common in both numerator and denominator, let us cancel the common term 4.
⇒u=12\Rightarrow u=\dfrac{1}{2}
But we have assumed u=sin⁡θu=\sin \theta .
∴sin⁡θ=12\therefore \sin \theta =\dfrac{1}{2}
Let us take sin⁡−1{{\sin }^{-1}} on both the sides of the expression, we get
sin⁡−1(sin⁡θ)=sin⁡−1(12){{\sin }^{-1}}\left( \sin \theta \right)={{\sin }^{-1}}\left( \dfrac{1}{2} \right)
Using the trigonometric property for inverse sin⁡−1(sin⁡θ)=θ{{\sin }^{-1}}\left( \sin \theta \right)=\theta , we get
⇒θ=sin⁡−1(12)\Rightarrow \theta ={{\sin }^{-1}}\left( \dfrac{1}{2} \right)
We know that sin⁡(π6)=12\sin \left( \dfrac{\pi }{6} \right)=\dfrac{1}{2}.
But from the trigonometric cycle, we understand that after an interval of 2π2\pi , the same value for sin function repeats.
Therefore, for π6+2nπ\dfrac{\pi }{6}+2n\pi , the value of sin function is 12\dfrac{1}{2}.
Hence, the values of θ\theta for the equation 4sin⁡2θ−4sin⁡θ+1=04{{\sin }^{2}}\theta -4\sin \theta +1=0 are π6+2nπ\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⁡θ=u\sin \theta =u and considering sin⁡θ\sin \theta being the variable in the quadratic equation.