Question

Solve the equation \[4{{\sin }^{2}}\theta -8\cos \theta +1=0\].

Answer 1

Hint: We will apply the formula of trigonometric identity which is given by \[{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1\]. Also, we will apply the hit and trial method where the factors are found out by substituting values to x. The formula \[{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1\] is applied to the question given in the question to simplify it and then solve it. We will apply it to the terms \[{{\sin }^{2}}\theta \].

Complete step-by-step answer:

We will consider the equation \[4{{\sin }^{2}}\theta -8\cos \theta +1=0\]. Here we will apply the formula of trigonometry, \[{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1\]. Therefore, we got a new expression, i.e. by applying \[{{\sin }^{2}}\theta =1-{{\cos }^{2}}\theta \] we have \[4\left( 1-{{\cos }^{2}}\theta \right)-8\cos \theta +1=0\].

Now, by arranging the equation in arranging order of degrees, we have,

\[\begin{align}

  & -4{{\cos }^{2}}\theta -8\cos \theta +4+1=0 \\

 & -4{{\cos }^{2}}\theta -8\cos \theta +5=0 \\

\end{align}\]

Now we will substitute \[\cos \theta =x\]. Therefore, we have \[{{\cos }^{2}}\theta =x\]. Thus we get a new equation written as,

\[-4{{x}^{2}}-8x+5=0...(2)\]

By using hit and trial method and substituting the value of x in equation (2) we get the factors of equation (2). First, we substitute the value of x = 0 in equation (2). Therefore we have,

\[-4{{\left( 0 \right)}^{2}}-8\left( 0 \right)+5=0-0+5\].

This value of x = 0 does not satisfy equation (2) as the equation does not result in 0. Now we will apply the value \[x=\dfrac{1}{2}\] and substitute it in equation (2).

\[\begin{align}
  & -4{{\left( \dfrac{1}{2} \right)}^{2}}-8\left( \dfrac{1}{2} \right)+5=-4\times \dfrac{1}{4}-\dfrac{8}{2}+5 \\

 & \dfrac{-4}{4}-\dfrac{8}{2}+5=-1-4+5 \\

 & -1-4+5=0 \\

\end{align}\]

This clearly implies that \[x=\dfrac{1}{2}\] is a factor of equation (2) or \[\left( x-\dfrac{1}{2} \right)\] is a factor.

Now we will divide \[\left( x-\dfrac{1}{2} \right)\] to equation (2). Thus we get,

\[x-\dfrac{1}{2}\overset{-4x-10}{\overline{\left){\begin{align}

  & -4{{x}^{2}}-8x+5 \\

 & -4{{x}^{2}}+2x \\

 & \overline{\left){\begin{align}

  & -10x+5 \\

 & \underline{-10+5} \\

 & \underline{000000} \\

\end{align}}\right.} \\

\end{align}}\right.}}\]

Clearly, we factorize the equation \[-4{{x}^{2}}-8x+5=0\] into \[\left( x-\dfrac{1}{2} \right)\left( -4x-10 \right)=0\].

\[\Rightarrow \left( x-\dfrac{1}{2} \right)\left( -4x-10 \right)=0\].

Now we have \[\left( x-\dfrac{1}{2} \right)=0\] or \[-4x-10=0\].

Consider, \[x-\dfrac{1}{2}=0\] or \[x=\dfrac{1}{2}\]. Since we know that \[x=\cos \theta \Rightarrow \cos \theta =\dfrac{1}{2}\].

As we know that \[\dfrac{1}{2}\] is the value of \[\cos \dfrac{\pi }{3}\].

\[\Rightarrow \cos \theta =\cos \dfrac{\pi }{3}\]

The value of cos is positive in the first and fourth quadrants. In quadrant 1, we have \[\cos \theta =\cos \dfrac{\pi }{3}\].

\[\begin{align}

  & \Rightarrow \cos \theta =\cos \dfrac{\pi }{3} \\

 & \Rightarrow \theta =\dfrac{\pi }{3} \\

\end{align}\]

or general solution is \[\theta =\dfrac{\pi }{3}+2n\pi \], where \[n=1\pm 2...\]

In the fourth quadrant, we have \[\cos \dfrac{\pi }{3}=\cos \left( 2\pi -\dfrac{\pi }{3} \right)\].

\[\begin{align}

  & \Rightarrow \cos \theta =\cos \left( 2\pi -\dfrac{\pi }{3} \right) \\

 & \Rightarrow \cos \theta =\cos \left( \dfrac{6\pi -\pi }{3} \right) \\

 & \Rightarrow \cos \theta =\cos \left( \dfrac{5\pi }{3} \right) \\

 & \Rightarrow \theta =\dfrac{5\pi }{3} \\

\end{align}\]

or general solution is \[\theta =\dfrac{5\pi }{3}+2n\pi \], where \[n=1\pm 2...\]

Now we will consider\[-4x=10\] or \[x=-\dfrac{5}{2}\].

Now we can write that, \[\cos \theta =-\dfrac{5}{2}\].

As we know that \[\cos \theta =-\dfrac{5}{2}\] results into \[\theta ={{\cos }^{-1}}\left(

-\dfrac{5}{2} \right)\].

\[\begin{align}

  & \cos \theta =\cos \theta \\

 & \theta =\theta \\

\end{align}\]

\[\theta =\theta +2n\pi \] where \[n=1\pm 2...\]

or the general solution is \[\theta ={{\cos }^{-1}}\left( \dfrac{-5}{2} \right)+2n\pi \].

Hence the solution of equation \[4{{\sin }^{2}}\theta -8\cos \theta +1=0\] is given by \[\theta =\dfrac{\pi }{3},\theta ={{\cos }^{-1}}\left( \dfrac{-5}{2} \right)\].

Note: Alternatively we could have factorized the equation \[4{{\sin }^{2}}\theta -8\cos \theta +1=0\] by factorization. Actually, we could have used factorization method in \[4{{\sin }^{2}}\theta -8\cos \theta +1=0\] by using the formula for square roots given by, \[x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\] where \[{{x}_{1}},{{x}_{2}}\] are roots of the equation \[a{{x}^{2}}+bx+c=0\].