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To solve $\sin x= 2\sin(x-30^\circ) \sin(40^{\circ})$ I expanded RHS by using $\sin(\alpha-\beta)=\sin\alpha\cos\beta-\sin\beta\cos\alpha$,

$$\sin x= 2\sin(40^{\circ})\times\left(\frac{\sqrt3}2\sin x- \frac12\cos x\right)$$ $$\left(\sqrt3 \sin(40^{\circ})-1\right)\sin x= \sin(40^{\circ})\cos x$$ $$\cot x= \sqrt 3- \csc(40^{\circ})$$

But from here I'm not sure how to continue.

I've also tried using $2\sin\alpha\sin\beta= \cos(\alpha-\beta)-\cos(\alpha+\beta)$ for the RHS which result in, $\sin x= \cos(x-70^\circ)-\cos(x+10^\circ)$ but don't know how to continue.

user
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Etemon
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    How about using arccot? – Taladris Jan 26 '24 at 04:52
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    @Taladris I checked the final answer is $x=80^{\circ}$ so I think it is possible to continue further. – Etemon Jan 26 '24 at 04:58
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    Not very satisfying: your computations show that $x=\alpha+n180^o$ with $\alpha=\cot^{-1}(\sqrt{3}-\csc(40^o))$ and $n$ can be any integer. Now, using $2\sin\alpha\sin\beta= \cos(\alpha-\beta)-\cos(\alpha+\beta) $ and $\cos(\theta)=\sin(90^o-\theta)$, it is easy to check that $80^o$ is indeed a solution. Therefore, $x=(80+180n)^o$. – Taladris Jan 26 '24 at 05:21

4 Answers4

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Since $\sin 40^\circ = \cos (90^\circ - 40^\circ) = \cos 50^\circ$, we have $$\begin{align} 0 &= -\sin x + 2 \sin (x - 30^\circ) \cos 50^\circ \\ &= -\sin x + \sin (x + 20^\circ) + \sin (x - 80^\circ) \\ &= 2 \sin 10^\circ \cos (x + 10^\circ) + \sin (x - 80^\circ) \\ &= 2 \sin 10^\circ \sin (90^\circ - (x + 10^\circ)) + \sin (x - 80^\circ) \\ &= 2 \sin 10^\circ \sin (80^\circ - x) + \sin (x - 80^\circ) \\ &= (1 - 2 \sin 10^\circ) \sin (x - 80^\circ). \end{align}$$ Since $1 - 2 \sin 10^\circ \ne 0$, this immediately implies $$x = 80^\circ + 180^\circ k$$ for any integer $k$.

heropup
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As an alternative, once checked that $80°$ is an answer let $x=y+80$ then

$$\sin x= 2\sin(x-30°) \sin 40°$$

$$\sin (y+80°)= 2\sin(y+50°) \sin 40°$$

$$\sin (y+80°)= 2\cos(40°-y) \sin 40°$$

$$\require{cancel}\sin y \cos 80°+\cancel{\cos y\sin 80°}= \cancel{\cos y\sin 80°} + 2\sin^2 40°\sin y$$

$$\sin y \cancel{(\cos 80°-2\sin^2 40°)}=0\iff \sin y =0 \iff y=k\pi$$

user
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We have that, by product to sum and sum to product identities

$$\sin x= 2\sin(x-30°) \sin 40°$$

$$\sin x= 2\cos(120°-x) \sin 40°$$

$$\sin x= \sin(160°-x) +\sin (80°-x)$$

$$\sin x+ \sin(x-160°) =\sin (80°-x)$$

$$2\sin (x-80°)\cos 80° =\sin (80°-x)$$

$$\sin (x-80°)\require{cancel}\cancel{(2\cos 80°+1)}=0 \iff \boxed{\sin (x-80°)=0}$$

user
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1

Let's see how the problem could have come into being ?

If $\sin A=2\sin A\cos2B\ \ \ \ (1)$ with $\cos2B\ne0,\sin A=0\iff A=n180^\circ$

$$(1)\iff\sin A=\sin(A-2B)+\sin(A+2B)$$

$$\iff\sin(A+2B)=\cdots=2\sin B\sin(90^\circ+A-B)$$

Comparing with $\sin x= 2\sin(x-30^\circ) \sin(40^{\circ}),$

Case $\#1:$

if $A+2B=x\iff A=x-2B$ and $90^\circ+A-B=x-30^\circ$

$B=120^\circ+A-x=120^\circ+(x-2B)-x\iff B=40^\circ$ (The current scenario)

Case $\#2:$

if $B=x-30^\circ$ and $A+2B=x\iff A=\cdots=60^\circ-x$

As $A=n180^\circ, x=?$

$90^\circ+A-B=\cdots=180^\circ-2x=?$

$\sin x=2\sin(x-30^\circ)\sin2x$ which is trivial, right?

lab bhattacharjee
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