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.
Hello, jo74!
Use the identities: .$\displaystyle \begin{array}{ccc}\sin2\theta &=& 2\sin\theta\cos\theta \\ \cos2\theta &=& 2\cos^2\!\theta - 1\end{array}$Solve the equation: .$\displaystyle
\sin2x - 1 \:= \:\cos 2x\;\;\text{ for }0^o < x < 360^o$
And we have: .$\displaystyle 2\sin x\cos x - 1 \;=\;2\cos^2\!x - 1 \quad\Rightarrow\quad 2\sin x\cos x - 2\cos^2\!x \;=\;0$
Factor: .$\displaystyle 2\cos x(\sin x - \cos x) \:=\:0 $
And we have two equations to solve:
. . $\displaystyle 2\cos x \:=\:0 \quad\Rightarrow\quad\boxed{x \:=\:90^o,\:270^o}$
. . $\displaystyle \sin x - \cos x \:=\:0 \quad\Rightarrow\quad \sin x \:=\:\cos x $
. . . . Divide by $\displaystyle \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 . . .
.
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.