Question

How do you solve \[2{{\cos }^{2}}x-\cos x=0\] from \[\left[ 0,360 \right]\]?

Answer 1

Hint: The solution of the equation for \[2{{\cos }^{2}}x-\cos x=0\] is the general solution of the roots of the given equation. So, we have to find the roots of the given equation and then we will get equations on\[\cos x\]. By the general equation of \[\cos x\] we can easily find the solutions for the roots, which will be the solutions for\[2{{\cos }^{2}}x-\cos x=0\].

Complete step by step answer:
For the given question we are given to solve the equation \[2{{\cos }^{2}}x-\cos x=0\] from the range \[\left[ 0,360 \right]\].
Let us consider the given equation as equation (1).
\[2{{\cos }^{2}}x-\cos x=0.........\left( 1 \right)\]
By pulling the \[\cos x\] term from LHS (left hand side) in the equation (1), we get
\[\Rightarrow \cos x\left( 2\cos x-1 \right)=0\]
Which means,
\[\cos x=0\text{ or cosx=-}\dfrac{1}{2}\]
Let us consider the above equations as equation (2) and equation (3) respectively.
\[\begin{align}
  & \cos x=0\text{ }...........\left( 2 \right) \\
 & \text{cosx=-}\dfrac{1}{2}...........\left( 3 \right) \\
\end{align}\]
As we know the general equation for \[\cos x\]is\[x=\dfrac{\left( 2n+1 \right)\pi }{2}\], where n is an integer.
Therefore let us consider the general equation as equation (4).
\[x=\dfrac{\left( 2n+1 \right)\pi }{2}.........\left( 4 \right)\]
Now by applying the general equation to equation (2), we get
\[x=\dfrac{\left( 2n+1 \right)\pi }{2}\]
Let us consider the above equation as equation (5), we get
\[x=\dfrac{\left( 2n+1 \right)\pi }{2}..........\left( 5 \right)\]
Now by applying the general equation to equation (3).
\[x=2n\pi \pm \dfrac{2\pi }{3}\]
Let us consider the above equation as equation (5).
\[x=2n\pi \pm \dfrac{2\pi }{3}..........\left( 5 \right)\]
Therefore from the equation (4) and equation (5) we can say that
General solution for \[2{{\cos }^{2}}x-\cos x=0\] is \[x=\dfrac{\left( 2n+1 \right)\pi }{2}\] or\[x=2n\pi \pm \dfrac{2\pi }{3}\].

Note: We can also find the roots of the given equation by the formula\[\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\]. Where ‘b’ is the coefficient of the term\[\cos x\], ‘a’ is the coefficient of \[{{\cos }^{2}}x\]and ‘c’ is the constant of the equation.