Matematické Fórum

stuart clark · 03. 11. 2011 16:41

solve for $x$

$1+\frac{1}{2\sin (30^0+x)}=\frac{\sin \frac{x}{2}}{\sin (\frac{x}{2}+60^0)}+\frac{\sqrt{3}}{2.\sin (\frac{x}{2}+60^0)}$

laszky · 16. 03. 2018 16:34 — Editoval laszky (16. 03. 2018 20:03)

Firstly we denote $s=\sin\left({\small \frac{x}{2}}\right)$ and $c=\cos\left({\small \frac{x}{2}}\right)$ and adjust all sines as follows

$\sin(30°+x) = \frac{1}{2}\cos x + \frac{\sqrt{3}}{2}\sin x = \frac{1}{2}\,(c^2-s^2) + \sqrt{3}\,cs$

$\sin\left(\frac{x}{2}+60°\right) = \frac{1}{2}\,s + \frac{\sqrt{3}}{2}\,c$ .

Next, we remove fractions from the equation and use $c^2=1-s^2$ . We obtain the equality

$c(2\sqrt{3}-6s-4\sqrt{3}s^2) = 4s^3-2\sqrt{3}s^2-6s+\sqrt{3}$

Further, we square the whole equation and employ the equality $c^2=1-s^2$ again. This yields

$-16\left(s+\frac{\sqrt{3}}{2}\right)^2\left(s-\frac{\sqrt{3}}{2}\right)\left(4s^3-3s+\frac{\sqrt{3}}{2}\right)=0$

While $s=\frac{\sqrt{3}}{2}$ is the solution of the original equation, $s=-\frac{\sqrt{3}}{2}$ is not.

In order to solve the cubic equation, we use the identity

$4\sin^3 y-3\sin y = \sin^3y + 3\sin y(\sin^2y -1) = \sin^3y - 3\sin y\cos^2 y =$
$= \sin y(\sin^2y-\cos^2y) -2\sin y\cos^2y=-\sin y\cos(2y)-\sin(2y)\cos y = -\sin(3y)$

Hence $\sin\left(\frac{3}{2}x\right)=\frac{\sqrt{3}}{2}$ .

The solutions of the given equation, thus, can be only of the form

$x_1=\frac{2}{3}\pi+4k\pi, \qquad x_2=\frac{4}{3}\pi+4k\pi, \qquad x_3=\frac{2}{9}\pi+\frac{4}{3}k\pi \qquad \mathrm{and}\qquad x_4=\frac{4}{9}\pi+\frac{4}{3}k\pi$ .

Plugging them into the original equation, we find out that only $x_1$ and $x_4$ solve it.

Matematické Fórum

#1 03. 11. 2011 16:41

Trigonometric equation

#2 16. 03. 2018 16:34 — Editoval laszky (16. 03. 2018 20:03)

Re: Trigonometric equation

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