1. ## trig identity

Solve for x in sinx tanx + tanx – 2sinx + cosx = 0 for 0≤ x ≤ 2π rad.?

2. Hello, yoman360!

The best I can do is reduce it to a cubic equation with no rational roots.

Solve for $x\!:\;\;\sin x\tan x + \tan x - 2\sin x + \cos x \:=\: 0\;\text{ for }0 \le x \le 2\pi$

We see that neither $\tfrac{\pi}{2}$ nor $\tfrac{3\pi}{2}$ is a root of the equation.

Divide by $\cos x\!:\;\;\frac{\sin x\tan x}{\cos x} + \frac{\tan x}{\cos x} - \frac{2\sin x}{\cos x} + \frac{\cos x}{\cos x} \:=\:\frac{0}{\cos x}$

. . . . . . . . . . . . . . $\tan^2\!x + \tan x\sec x - 2\tan x + 1 \:=\:0$

. . . . . . . . . . . . . . . . . . . . . $\tan^2\!x - 2\tan x + 1 \;=\;-\tan x\sec x$

. . . . . . . . . . . . . . . . . . . . . . . . . . $(\tan x - 1)^2 \;=\;-\tan x\sqrt{\tan^2\!x+1}$

Square both sides: . . . . . . . . . . . . $(\tan x - 1)^4 \;=\;\tan^2\!x(\tan^2\!x + 1)$

. . . . . . . . . $\tan^4\!x - 4\tan^3\!x + 6\tan^2\!x - 4\tan x + 1 \;=\;\tan^4\!x + \tan^2\!x$

And we have: . . . . $4\tan^3\!x - 5\tan^2\!x + 4\tan x - 1 \:=\;0$

. . This cubic equation has no rational roots.

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I realize that the roots need not be rational.

For example, if $x = \tfrac{\pi}{6}\;(30^o)$, then $\tan\tfrac{\pi}{6} = \tfrac{1}{\sqrt{3}}$ could be a root . . . (It isn't.)

And if $x = \tfrac{\pi}{12}\;(15^o)$, then $\tan\tfrac{\pi}{12} = 2-\sqrt{3}$ could be a root . . . (It isn't.)

At this point, I surrendered . . .

3. Hello,

Let's multiply by $\cos(x)$. This gives :

$\sin^2(x)+\sin(x)-2\sin(x)\cos(x)+\cos^2(x)=0$

but we know that $\cos^2(x)+\sin^2(x)=1$ and $\sin(2x)=2\sin(x)\cos(x)$

so the equation is now

$\sin(x)-\sin(2x)+1=0$

Perhaps it has a better form this way...

4. For what it is worth, pursuing Moo's initial stages, by iteration the solution is 1.1131Pi or 3.49683 radians. There appears to be only the one real solution.