MHB Can You Solve This Trigonometric Equation for $x$?

AI Thread Summary
The discussion revolves around solving the trigonometric equation $2\sin(x+30^\circ)\sin 16^\circ \sin 76^\circ=\sin 2028^\circ \sin 210^\circ$ for $0 < x < 180^\circ$. Participants engage in deriving solutions and simplifying the equation, with some noting the importance of understanding the sine function's periodicity. A solution was identified that had initially been overlooked, highlighting the collaborative nature of problem-solving in the forum. The conversation emphasizes the significance of careful analysis in trigonometric equations. Overall, the thread showcases effective mathematical discourse and solution-finding.
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Solve for $x$ such that $2\sin(x+30^\circ)\sin 16^\circ \sin 76^\circ=\sin 2028^\circ \sin 210^\circ$ for $0\lt x \lt 180^\circ$.
 
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anemone said:
Solve for $x$ such that $2\sin(x+30^\circ)\sin 16^\circ \sin 76^\circ=\sin 2028^\circ \sin 210^\circ$ for $0\lt x \lt 180^\circ$.

$$\sin2028^\circ\sin210^\circ=-\dfrac{\sin228^\circ}{2}=\dfrac{\sin48^\circ}{2}$$

$$\sin48^\circ=4\sin(x+30^\circ)\sin16^\circ\sin76^\circ$$

$$3\sin16^\circ-4\sin^316^\circ=4\sin(x+30^\circ)\sin16^\circ\sin76^\circ$$

$$1+2\cos32^\circ=4\sin(x+30^\circ)\sin76^\circ$$

$$1+2\cos32^\circ=2(\cos(x-46^\circ)-\cos(x+106^\circ))$$

$$\text{By inspection, }x=14^\circ$$
 
anemone said:
Solve for $x$ such that $2\sin(x+30^\circ)\sin 16^\circ \sin 76^\circ=\sin 2028^\circ \sin 210^\circ$ for $0\lt x \lt 180^\circ$.

$2\sin(x+30^\circ)\sin 16^\circ \sin 76^\circ=\sin 2028^\circ \sin 210^\circ$
= $\sin (360^\circ * 5 + 180^\circ + 48^\circ) \sin (180^\circ +30^\circ)$
= $(-\sin\, 48^\circ) (-\sin \,30^\circ)$
= $\dfrac{\sin\,48^\circ}{2}$
= $\dfrac{\sin\, 3*16^\circ}{2}$
= $\dfrac{3\sin\,16^\circ-4\sin ^3 16^\circ}{2}$
hence
$2\sin(x+30^\circ)\sin 16^\circ \sin 76^\circ = \sin\,16^\circ \dfrac{3-4\sin ^2 16^\circ}{2}$
or
$4\sin(x+30^\circ)\sin 76^\circ = 3-4\sin ^2 16^\circ$
= $ 1 + 2(1-2\sin^2 16^\circ)$
= $ 1+ 2 \cos\,32^\circ$
= $(2(\cos\,60^\circ +2 \cos\,32^\circ)$
= $ 2 (2 \cos\,46^\circ \cos\,14^\circ)$
= $ 4 \cos\,46^\circ \sin \,76^\circ$
hence
$\sin(x+30^\circ)=\cos\,46^\circ=\sin\,44^\circ$
or $x=14^\circ$

edit there is a solution I missed based on comment below
in the 2nd quadrant
$\sin(x+30^\circ)=\sin\,136^\circ$
so $x= 106^\circ$
 
Last edited:
Thanks both for participating and the solution...but...

Are you certain you haven't missed any solution? (Mmm)
 
anemone said:
Thanks both for participating and the solution...but...

Are you certain you haven't missed any solution? (Mmm)
I missed the solution $x + 30^\circ = 136^\circ$ or $x = 106^\circ$
 
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