Question

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

Answer 1

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.

& \cos x=0\text{ }...........\left( 2 \right) \\\ & \text{cosx=-}\dfrac{1}{2}...........\left( 3 \right) \\\ \end{aligned}$$ As we know the general equation for $$\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.