1. Solving Trig Equations

Incorporate the use of the double angle formulas identities and other identities to solve for the following on $0 \leq \theta \leq 360$

$tanx = sin2x$

My Attempt:
$\frac{sinx}{cosx} = 2sinxcosx$

$sinx = 2sinx$

$sinx - 2sinx = 0$

$-sinx = 0$

$x =$ 0 and 180 and 360

0 and 45 and 135 and 180 and 225 and 315 and 360

What am I doing wrong?

2. Hello, Macleef!

A silly mistake . . . don't kick yourself too hard.

Solve for $0 \leq \theta \leq 360$

$\tan x \:= \:\sin2x$

My Attempt:

$\frac{\sin x}{\cos x} \:= \:2\sin x\cos x$

$\sin x \:= \:2\sin x$ . . . . no
The cosines don't cancel . . .

We have: . $\sin x \:=\:2\sin x\cos^2\!x$

. . $\sin x - 2\sin x\cos^2\!x \:=\:0$

Factor: . $\sin x(1 - 2\cos^2\!x) \:=\:0$

Then: . $\sin x \:=\:0\quad\Rightarrow\quad \boxed{x \:=\:0^o,\:180^o,\:360^o}$

And: . $1 - 2\cos^2\!x \:=\:0\quad\Rightarrow\quad \cos^2\!x \:=\:\frac{1}{2}$
. . . . $\cos x \:=\:\pm\frac{1}{\sqrt{2}}\quad\Rightarrow\quad \boxed{x \;=\;45^o,\:135^o,\:225^o,\:315^o}$

3. Originally Posted by Macleef
Incorporate the use of the double angle formulas identities and other identities to solve for the following on $0 \leq \theta \leq 360$

$tanx = sin2x$

My Attempt:
$\frac{sinx}{cosx} = 2sinxcosx$

$sinx = 2sinx$

$sinx - 2sinx = 0$

$-sinx = 0$

$x =$ 0 and 180 and 360

0 and 45 and 135 and 180 and 225 and 315 and 360

What am I doing wrong?
What did you do wrong?
From sinX /cosX = 2sinXcosX, how did you get sinX = 2sinX? What happened to the two cosX's?

You want me to solve the Problem?

Bring them all to the lefthand side,
sinX/cosX -2sinXcosX = 0
sinX(1/cosX -2cosX) = 0

sinX = 0
X = arcsin(0) = 0, 180, 360 degrees ---------***

1/cosX -2cosX = 0
Clear the fraction, multiply both sides by cosX,
1 -2cos^2(X) = 0
1 = 2cos^2(X)
1/2 = cos^2(X)
cosX = +,-sqrt(1/2)
cosX = +,-1/sqrt(2)
X = arccos(+,-1/sqrt(2))
X = 45, 135, 225, 315 degrees. ------***

Therefore, X = 0, 45, 135, 180, 225, 315, 360 degrees.