Question

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)?

Answer 1

Answer:

1. C

2. B

3. A, E

4. B

Step-by-step explanation:

Answer 2

$sinx=cosx-1 \\ \\ sinx-cosx=-1$

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

$- \sqrt{2}cos( \frac{ \pi }{4}+x)=-1 \\ \\ cos( \frac{ \pi }{4}+x)= \frac{1}{ \sqrt{2} }$

There are two solutions to this equation:

1)
$\frac{ \pi }{4}+x= \frac{ \pi }{4} \\ \\ x=0$

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

2) 
$\frac{ \pi }{4}+x= \frac{7 \pi }{4} \\ \\ x= \frac{3 \pi }{2}$

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