Question

If $2{\sin ^2}x - \cos x = 1$ how do you solve for $x$?

Answer 1

Hint:We, are going to first convert ${\sin ^2}x$ in terms of $\cos x$, such that we have the whole quadratic equation in terms of $\cos x$ and then we solve the equation to get its solutions and hence get the values of $x$.

Complete step by step answer:
We are given the equation, $2{\sin ^2}x - \cos x = 1$. Since we know that ${\sin ^2}x + {\cos ^2}x = 1$, using this we are going to convert ${\sin ^2}x$ in terms of $\cos x$ which is ${\sin ^2}x = 1 - {\cos ^2}x$, we are going to substitute in the equation. We get
$2(1 - {\cos ^2}x) - \cos x - 1 = 0 \\
\Rightarrow 2 - 2{\cos ^2}x - \cos x - 1 = 0 \\ $
Now, we have a quadratic equation in terms of $\cos x$, we are going to solve it.It is of the form $(a - b + c = 0)$, we can use a shortcut to get the solutions.We will equate the roots to the coefficient of the given quadratic equations, such that we can solve the solutions. So,
$\cos x = 1\,\,and\,\,\cos x = \dfrac{c}{a} = - \dfrac{1}{2}$
We are going to solve the above two equations, then we get the solutions
\[\cos x = 1\]
On further simplification, we get \[arc\,x = 0\], also, we can also get,
\[x = 2\pi \]
Then we have
$\cos x = - \dfrac{1}{2}$
Which further simplify to get the value of x, so
$arc = \pm \dfrac{{2\pi }}{3}$

Hence, we have solved the given equations and found the solutions for it. They are
$0,2\pi , \pm \dfrac{{2\pi }}{3}$.

Note:We should be careful while solving, as we have to not only consider one angle, we should try all possible angles where we all can get the value of. We have to make sure and check the equation solving again for any minor mistakes like missed variables or interchanged signs.