# Solving a trig equation?

How would I solve the following equation? $$1+ \sin (x)=2 \cos(x)$$ I am having difficult with it.

-

We have $1 + \sin(x) = 2 \cos(x)$. Recall that $1 - \sin^2(x) = \cos^2(x)$. Hence, we get that $$(1 + \sin(x))(1- \sin(x)) = \cos^2(x)$$ i.e. $$2 \cos(x) (1 - \sin(x)) = \cos^2(x)$$ If $\cos(x) \neq 0$, then we get that $$1 - \sin(x) = \dfrac{\cos(x)}2$$ Hence, have \begin{align} 1 + \sin(x) & = 2\cos(x)\\ 1 - \sin(x) & = \dfrac{\cos(x)}2 \end{align} This gives us that $2 = \dfrac52 \cos(x) \implies \cos(x) = \dfrac45$. This gives us $\sin(x) = \dfrac35$. Hence, we get one possible solution as $$x = 2n \pi + \theta$$ where $\sin(\theta) = \dfrac35$ with $0 < \theta < \dfrac{\pi}2$.

If $\cos(x) = 0$, then we need $\sin(x) = -1$. Hence, $x = 2n\pi + \dfrac{3 \pi}2$.

-
,according to my method, $x=\arccos \frac45=\arcsin\frac45$ is another solution. –  lab bhattacharjee Dec 15 '12 at 5:19
@labbhattacharjee Try plugging your answer in... $1 + \frac{4}{5} \neq 2 \cdot \frac{4}{5}$, so it looks like something went wrong. –  andybenji Dec 15 '12 at 5:51
@labbhattacharjee Thanks. Yes. the other solution is $x = \arcsin(3/5) = \arccos(4/5)$. –  user17762 Dec 15 '12 at 6:03
@andybenji, sorry there was terrible typo--it should be $\arcsin\frac35=\arccos\frac45$ as Marvis has noted. –  lab bhattacharjee Dec 15 '12 at 6:58

Hint: Square both sides, and express $\cos^2 x$ in terms of $\sin^2 x$.

You will get a quadratic equation in $\sin^2 x$. After finding the solutions of this equation, make sure to substitute into the original equation to check whether they are indeed roots of the original equation. The squaring process could produce "spurious" roots. (It happens not to, but checking is cheap.)

-
I have a quick question I squared both sides and got. 1+2sin(x)+sin^2(x)=4cos^2(x) so I made 4cos^2(x) into 4-sin^2(x) Is this correct. –  Fernando Martinez Dec 15 '12 at 4:32
It would be $4-4\sin^2 x$. The equation then becomes $5\sin^2 x+2\sin x-3=0$. You can then use the Quadratic Formula or factor as $(5\sin x-3)(\sin x+1)$. –  André Nicolas Dec 15 '12 at 4:48
Oh thanks very much. –  Fernando Martinez Dec 15 '12 at 4:53

This may be the hard way, but if you let $q=\tan(x/2)$, then $\sin x=2q/(1+q^2)$, and $\cos x=(1-q^2)/(1+q^2)$.

-

Let me take a stab at this. I had to double check a few identities. First, we use the identities $\cos(\frac\pi 2-x)=\sin x$ and $\sin(\frac\pi 2-x)=\cos x$.

$$1+\cos(\frac\pi 2-x)=2\sin(\frac\pi 2-x)$$ $$\frac{1+\cos(\frac\pi 2-x)}{\sin(\frac\pi 2-x)}=2$$

Next, we use the half-angle identity $\tan\frac\theta 2=\frac{\sin\theta}{1+\cos\theta}$.

$$\cot(\frac\pi 4-\frac x2)=2$$ $$\tan(\frac\pi 4-\frac x2)=\frac12$$

From here, take the inverse tangent and the rest is algebra.

-
$\cos\left(\frac\pi4-\frac x2\right)$ can also be $0$ –  lab bhattacharjee Dec 15 '12 at 9:38
@labbhattacharjee I see how I might have lost some roots by not making sure $\cos x$, or $\sin(\frac\pi 2-x)$ if you prefer, was 0 before dividing by it. Did I lose roots in any other step somehow? –  Mike Dec 15 '12 at 18:22

We can try to avoid squaring to avoid tests for extraneous root. $$2\cos x-\sin x=1$$

Putting $r\cos\theta=2,r\sin\theta=1$ where $r>0$

Squaring and adding we get $r^2=2^2+1^2\implies r=\sqrt5$

On division, $\tan \theta=\frac12$ where $0<\theta<\frac\pi2$ as $\sin\theta,\cos\theta>0$

we get $$\cos\theta\cos x-\sin\theta\sin x=\sin\theta$$(cancelling $r$ in either sides )

or, $\cos(x+\theta)=\cos(\frac\pi2-\theta)$

So, $$x+\theta=2n\pi\pm\left(\frac\pi2-\theta\right)$$ where $n$ is any integer.

Taking $'-'$ sign, $$x+\theta=2n\pi-\left(\frac\pi2-\theta\right)\implies x=2n\pi-\frac\pi2$$

Taking $'+'$ sign, $$x+\theta=2n\pi+\left(\frac\pi2-\theta\right)\implies x=2n\pi+\frac\pi2-2\theta=2n\pi+\frac\pi2-2\arctan \frac12$$

Now, we know $\cos2y=\frac{1-\tan^2y}{1+\tan^2y},\sin2y=\frac{2\tan y}{1+\tan^2y},\tan2y=\frac{2\tan y}{1-\tan^2y}$

If $\arctan \frac12=y,\tan y=\frac 12$

$$\cos2y=\frac35,\sin2y=\frac45,\tan2y=\frac43$$

So, $$2y=2\arctan \frac12=\arccos\frac35=\arcsin\frac45=\arctan \frac43$$

So, $$x=2n\pi+\frac\pi2-\arccos\frac35=2n\pi+\arcsin\frac35$$ as $\arcsin z+\arccos z=\frac12$ for $-1\le z\le1$

Similarly, $$x=2n\pi+\arccos\frac45=2n\pi+\arccot\frac43$$

-
It is not clear we should avoid squaring, since checking is cheap. A good reason to avoid squaring is if degree is increased beyond the comfortable. But degree $2$ is comfortable. –  André Nicolas Dec 15 '12 at 5:48
@AndréNicolas, squaring is easing the calculation a lot for $A\cos x+B\sin x=C$ w/o introducing much intricacies. There is no extraneous root as such,right?. We just need to identify the quadrant of $x$ by getting the sign of $\cos x$ from the given equation, if we have the quadratic equation in $\sin x$ or vice versa. Is $\arccot$ not a valid latex? –  lab bhattacharjee Dec 15 '12 at 9:44
@labbhattacharjee Apparently not. Try "\mathrm{arccot}" –  Thomas Dec 15 '12 at 10:32