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

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

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.

Answer 2

An equation is formed of two equal expressions. There are 3 solutions to the given trigonometric equation.

What is an equation?

An equation is formed when two equal expressions are equated together with the help of an equal sign '='.

The trigonometric equation that is given to us is,

$\rm Sin(x) = Cos(x) - 1$

the equation when simplified it can be written as,

$\rm Sin(x) = Cos(x) - 1\\\\Sin(x)-Cos(x) = -1$

Now, if we use the algebraic expression of sin(x)-cos(x), then the equation can be written as,

$Sin(x)-Cos(x) = -1\\\\-\sqrt2\ Cos(\dfrac{\pi}{4}+x)=-1\\\\\sqrt2\ Cos(\dfrac{\pi}{4}+x)=1\\\\Cos(\dfrac{\pi}{4}+x)=\dfrac{1}{\sqrt2}$

There are two solutions to the above equation, therefore, the solutions of the equation are,

1.

  $\dfrac{\pi}{4} + x = \dfrac{\pi}{4} \\\\x = 0$

Thus, one of the solutions of the given equation will be when x is 0, and the value of x will become zero when x is equal to 0 or 2π.

2.

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

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

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