Please can you show me how to solve
sin2x-1=cos2x
i have tried various methods but am getting nowhere
this is my first post so thanks in advance for the help

2. Originally Posted by chipper
Please can you show me how to solve
sin2x-1=cos2x
i have tried various methods but am getting nowhere
this is my first post so thanks in advance for the help
just use the double angle formulas for sine and cosine:

$\sin 2x = 2 \sin x \cos x$ and $\cos 2x = \cos^2 x - \sin^2 x$

3. i have tried these but think i am missing something, i can get to the point
2sinxcosx-1=2cos^2x-1=0
2sinxcosx-1-2cos^2x+1=0
2sinxcosx-2cos^2x=0
but then am abit lost how to solve, can't find a way to factorise it, or have i done something wrong

4. Originally Posted by chipper
i have tried these but think i am missing something, i can get to the point
2sinxcosx-1=2cos^2x-1=0
2sinxcosx-1-2cos^2x+1=0
2sinxcosx-2cos^2x=0
but then am abit lost how to solve, can't find a way to factorise it, or have i done something wrong
factor out the cosine

you get: $\cos x ( \sin x - \cos x ) = 0$

$\Rightarrow \cos x = 0$ or $\sin x - \cos x = 0$

can you continue?

5. if i continue i get
cosx=0
x=arccos 0
x=pi/2 (1.57)
but am unsure how to do the next bit

6. Originally Posted by chipper
if i continue i get
cosx=0
x=arccos 0
x=pi/2 (1.57)
but am unsure how to do the next bit
$\cos x = 0 \implies x = \frac {\pi}2 + n \pi$ for $n$ an integer

$\sin x - \cos x = 0$

$\Rightarrow \sin x = \cos x$

$\Rightarrow x = \frac {\pi}4 + k \pi$ for $k$ an integer

if you want $0 \le x \le 2 \pi$, then your solutions are: $x = \frac {\pi}2, \frac {3 \pi}2, \frac {\pi}4, \mbox { and } \frac {5 \pi}4$

be sure to check all these solutions in the original equation to make sure none are erroneous

7. thanks so much for your help