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Hi!

To solve this inequation (which we'll transform into an equation for now), we need to isolate x:

$$-sin\frac{\pi}{2}(x-1)+9 =8$$

We start by eliminating the constant 9:

$$-sin\frac{\pi}{2}(x-1)+9-9 =8-9$$

$$-sin\frac{\pi}{2}(x-1)=-1$$

Next, we eliminate the negative signs on each side of the equation:

$$sin\frac{\pi}{2}(x-1)=1$$

Now we need to use our unit circle to find the angle for which the sin is 1:

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We thus find that the angle \(\frac{\pi}{2}\) has a sine of 1. Therefore, our angle \(\frac{\pi}{2}(x-1)\) must be equivalent to this angle, which gives us this equation:

$$\frac{\pi}{2}(x-1)=\frac{\pi}{2}$$

All that remains is to solve the equation:

$$\frac{\pi}{2}(x-1) \div \frac{\pi}{2}=\frac{\pi}{2} \div \frac{\pi}{2}$$

$$(x-1)=\frac{\pi}{2} \div \frac{\pi}{2}$$

$$x-1=1$$

$$x-1+1=1+1$$

$$x=2$$

In other words, we find that the function \(-sin\frac{\pi}{2}(x-1)+9\) passes through y=8 at x=2.

You must then identify the period of the function, then express your answers in the following form:

$$x + pn, n ∈ ℤ$$

where \(x\) is the answer obtained when solving (we found \(x= 2\)), and where \(p\) is the period that you will have to find.

\(n\) is simply an integer that allows us to find any solution, because we remember that since it is a periodic function, there is an infinite number of solutions. \(x\) repeats itself infinitely in the following and previous cycles, so by choosing a certain integer \(n\) of our choice, we will obtain the answer for a certain given cycle.

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Finally, before giving your final answer, it's finally time to analyze the inequality sign we had. We were looking for the values ​​of x for which the function is less than or equal to y=8. Since y=8 is the minimum of the function, then we can't be less than that, but we can be equal. So, the final answer will be:

$$2 + pn, n ∈ ℤ$$

I'll let you find out p ;)

You should consult the following sheet, it presents the steps to follow to solve sine equations :)

Solving a Sine Equation or Inequality | Secondaire | Alloprof

I hope this helps you! :)