# Solve $2 \cos^2 x+ \sin x=1$ for all possible $x$

$$2\cos^2 x+\sin x=1$$

$$\Rightarrow 2(1-\sin^2 x)+\sin x=1$$

$$\Rightarrow 2-2 \sin^2 x+\sin x=1$$

$$\Rightarrow 0=2 \sin^2 x- \sin x-1$$

And so:

$$0 = (2 \sin x+1)(\sin x-1)$$

So we have to find the solutions of each of these factors separately:

$$2 \sin x+1=0$$

$$\Rightarrow \sin x=\frac{-1}{2}$$

and so $$x=\frac{7\pi}{6},\frac{11\pi}{6}$$

Solving for the other factor,

$$\sin x-1=0 \Rightarrow \sin x=1$$

And so $$x=\frac{\pi}{2}$$

Now we have found all our base solutions, and so ALL the solutions can be written as so:

$$x= \frac{7\pi}{6} + 2\pi k,\frac{11\pi}{6} + 2\pi k, \frac{\pi}{2} + 2\pi k$$

• And the question is.... ?? Nov 21 '19 at 23:36
• It's tagged with proof verification. The solution provided in the question is correct. Nov 21 '19 at 23:39
• @RobertShore OK, thx. Then, when an answer should be set? When it corrects the question? Nov 21 '19 at 23:41
• I'll provide an answer rather than a comment if the answer provided is wrong in some material way or if there's an alternative solution that provides additional insight. Nov 21 '19 at 23:43

Yes your solution is very nice and correct, as a slightly different alternative

$$2\cos^2(x)+\sin (x)=1 \iff 2(1-\sin x)(1+\sin x)+\sin x-1=0$$

$$\iff (\sin x-1)(-2-2\sin x)+(\sin x-1)=0 \iff (\sin x-1)(-1-2\sin x)=0$$

which indeed leads to the same solutions, or also from here by $$t=\sin x$$

$$2-2 \sin^2 x+\sin x=1 \iff 2t^2-t-1=0$$

$$t_{1,2}=\frac{1\pm \sqrt{9}}{4}=1, -\frac12$$

Your method's fine, the answer's right. The only improvement I can suggest is to make the definition of "base solution" clear upfront. Each "and so" acts as if a specific value of $$\sin x$$ has finitely many solutions rather than finitely many per period, so before you obtain them you should mention a restriction to $$[0,\,2\pi)$$ and then extend to $$\Bbb R$$ at the end.

Other way is used identities double angle and sum-product

$$\begin{eqnarray*} 2\cos^2x+\sin x& = & 1 \\ 2\cos^2x-1+\sin x& = & 0\\ \cos(2x)+\sin x& = & 0\\ \cos(2x)+\cos\left(\frac{\pi}{2}-x\right) & = & 0\\ 2\cos\left(\frac{x}{2}+\frac{\pi}{4}\right)\cos\left(\frac{3x}{2}-\frac{\pi}{4}\right) & = & 0\\ \end{eqnarray*}$$

$$2\cos^2(x)+\sin(x)=1$$

Using $$2\cos^2(x)-1=\cos(2x)$$ we have $$\cos(2x)=-\sin(x)=\cos(\pi/2+x).$$

Hence $$2x=\pm(\pi/2+x)+2n\pi$$, $$n\in\mathbb{Z}$$ and $$x=\pi/2+2n\pi$$ or $$x=-\pi/6+2n\pi/3.$$ The second expression can be re-written as $$-\pi/6\pm2\pi/3+2k\pi$$ or $$-\pi/6+2k\pi$$ giving the three solutions

\begin{align} x&=\pi/2+2n\pi\\x&=-\pi/6+2k\pi=11\pi/6+2k\pi\\x&=-5\pi/6+2k\pi=7\pi/6+2k\pi\\ \end{align}

Note that one of the solutions from the second expression, $$-\pi/6+2\pi/3+2k\pi=\pi/2+2k\pi$$ is absorbed into the first of the three solutions. This happens because this is a double root, tangential to the x-axis - see plot from Wolfrom Alpha: 