How many true solutions does the equation sinx=cosx-1 have over the interval 0 is less than or equal to x which is less than or equal to 2(pi)?

[tex]sinx=cosx-1 \\ \\ sinx-cosx=-1[/tex]

Using the identity: [tex]sinx-cosx=- \sqrt{2}cos( \frac{ \pi }{4}+x) [/tex], we get:

[tex]- \sqrt{2}cos( \frac{ \pi }{4}+x)=-1 \\ \\ cos( \frac{ \pi }{4}+x)= \frac{1}{ \sqrt{2} } \\ \\ [/tex]

There are two solutions to this equation:

1)
[tex] \frac{ \pi }{4}+x= \frac{ \pi }{4} \\ \\ x=0 [/tex]

Since the period of cosine is 2π, so 0 + 2π = 2π will also be a solution to the given equation

2) 
[tex] \frac{ \pi }{4}+x= \frac{7 \pi }{4} \\ \\ x= \frac{3 \pi }{2} [/tex]

Therefore, there are 3 solutions to the given trigonometric equation.