Can You Solve This Trigonometric Equation for $x$?

Click For Summary

Discussion Overview

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 $x$ within the interval $0 < x < 180^\circ$. The focus is on exploring potential solutions and methods for solving this equation.

Discussion Character

  • Mathematical reasoning

Main Points Raised

  • Multiple participants present the same equation to solve for $x$, indicating a focus on finding solutions.
  • One participant acknowledges a missed solution based on a comment from another participant, suggesting that there may be additional insights or methods to consider.
  • There are expressions of gratitude towards participants for their contributions, indicating a collaborative atmosphere.

Areas of Agreement / Disagreement

The discussion does not appear to reach a consensus on the solution, as participants express differing views and acknowledge missed solutions without resolving the equation definitively.

Contextual Notes

Some assumptions regarding the values of the trigonometric functions involved may be implicit, and the discussion does not clarify all mathematical steps necessary to arrive at a solution.

anemone
Gold Member
MHB
POTW Director
Messages
3,851
Reaction score
115
Solve for $x$ such that $2\sin(x+30^\circ)\sin 16^\circ \sin 76^\circ=\sin 2028^\circ \sin 210^\circ$ for $0\lt x \lt 180^\circ$.
 
Mathematics news on Phys.org
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 $x = 106^\circ$
 

Similar threads

Replies
8
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
Replies
3
Views
2K
  • · Replies 4 ·
Replies
4
Views
4K
  • · Replies 15 ·
Replies
15
Views
3K
  • · Replies 12 ·
Replies
12
Views
3K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 28 ·
Replies
28
Views
3K