1. ## Solve the equation

Can someone help me please. I don't know which way to go with this.

Solve the equation

sin2x - 1 = cos 2x

giving answers for values of x between 0 and 360 degrees.

2. use this for

$sin2(x)=2sin(x)cos(x)$

and

$cos2(x)=2cos^2(x)-1$

sub these and find the solution

3. Hello, jo74!

Solve the equation: . $
\sin2x - 1 \:= \:\cos 2x\;\;\text{ for }0^o < x < 360^o$
Use the identities: . $\begin{array}{ccc}\sin2\theta &=& 2\sin\theta\cos\theta \\ \cos2\theta &=& 2\cos^2\!\theta - 1\end{array}$

And we have: . $2\sin x\cos x - 1 \;=\;2\cos^2\!x - 1 \quad\Rightarrow\quad 2\sin x\cos x - 2\cos^2\!x \;=\;0$

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

And we have two equations to solve:

. . $2\cos x \:=\:0 \quad\Rightarrow\quad\boxed{x \:=\:90^o,\:270^o}$

. . $\sin x - \cos x \:=\:0 \quad\Rightarrow\quad \sin x \:=\:\cos x$
. . . . Divide by $\cos x\!:\;\;\frac{\sin x}{\cos x} \:=\:1 \quad\Rightarrow\quad \tan x \:=\:1 \quad\Rightarrow\quad\boxed{x \:=\:45^o,\:225^o}$

Edit: Ah . . . Amer already explained the game plan . . .
.

4. Originally Posted by jo74
Can someone help me please. I don't know which way to go with this.

Solve the equation

sin2x - 1 = cos 2x

giving answers for values of x between 0 and 360 degrees.
Here's a proof of the double angle formula for the sine and cosine functions so that you can better understand them.

sin 2α = sin (α + α)
sin 2α = sin α cos α + cos α sin α
sin 2α = 2 sin α cos α

So, there's two ways to substitute here. The question you have to ask yourself is which one will be more efficacious.