Solve this. Show work as detailed as possible.
$$\sqrt{\cos x} = 2\cos x-1$$
My work:
\begin{align*} 2\cos x & = \sqrt{\cos x}+1\\ \cos x & = \frac{\sqrt{\cos x}+1}{2}\\ x & = \cos^{-1}\left(\frac{\sqrt{\cos x}+1}{2}\right) \end{align*}
Is that correct ?????
$\endgroup$ 35 Answers
$\begingroup$$$\sqrt{\cos x}=2\cos x-1$$
$$\iff 2\left(\sqrt{\cos x}-1\right)\left(\sqrt{\cos x}+\frac{1}{2}\right)=0$$
$\sqrt{\cos x}\ge 0$, so $$\sqrt{\cos x}=1\iff \cos x=1\iff x=2\pi n$$
for some $n\in\Bbb Z$.
$\endgroup$ 4 $\begingroup$Hint
Square both sides first, and then solve the quadratic equation to find what $\cos x$ is.
Hint 2
$$\cos x = (2\cos x -1)^2= 4\cos^2x -4\cos x+1$$ $$4t^2-5t+1=0\quad \longrightarrow \quad \cos x = t = ?$$ $$x=?$$
$\endgroup$ 8 $\begingroup$First square both sides, then we find $$\cos x = (2\cos x -1)^2= 4\cos^2x -4\cos x+1.$$ Bringing everything to the same side gives $$4\cos^2x -5\cos x+1 = 0.$$ This can be solved like a normal quadratic equation, which gives $$\cos x = \frac{5 \pm \sqrt{(-5)^2-4\cdot4}}{2\cdot 4}.$$ This then gives $$\cos x = \frac{5\pm \sqrt{9}}{8}= \frac{5\pm 3}{8}.$$
Thus we find $$\cos x = 1, \qquad \text{or} \qquad \cos x = \frac{1}{4}.$$
Solving for $x$ gives $$x = 0, \qquad \text{or} \qquad x = \arccos{\frac{1}{4}}.$$
However, if we now check our original results, we see that only $x =0$ is a correct solution. Note that this is a solution on the interval $[0,2\pi]$. One can extend this easily to all of $\mathbb{R}$.
$\endgroup$ 7 $\begingroup$Take $\cos x=t$ and solve the equation for $t$. Then, undo the change and make sure the solution make sense (e.g. if $t=2$, then theres is no valid $x$).
$\endgroup$ 1 $\begingroup$Let $\cos x = c$
\begin{align*} c & = 4c^2 - 4c + 1\\ 0 & = 4c^2 - 5c + 1\\ 0 & = (4c-1)(c-1) \end{align*}
$c = 1/4$ or $c = 1$
Check and see that only $c=1$ works so
$ x= 2n \pi $
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