How would I solve the following equation? $ 1+ \sin (x)=2 \cos(x) $ I am having difficult with it.
Solving a trig equation $1+ \sin (x)=2 \cos(x)$?
5 Answers
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, $x = 2n\pi + \dfrac{3 \pi}2$.
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0@andybenji, sorry there was terrible typo--it should be $\arcsin\frac35=\arccos\frac45$ as Marvis has noted. – 2012-12-15
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.)
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0It 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 $(5\sin x-3)(\sin x+1)$. – 2012-12-15
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.
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0@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? – 2012-12-15
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, $$x=2n\pi+\arccos\frac45=2n\pi+\operatorname{arccot}\frac43$$
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0@labbhattacharjee Apparently not. Try "\mathrm{arccot}" – 2012-12-15