====== Question 6, Exercise 9.1 ======
Solutions of Question 6 of Exercise 9.1 of Unit 09: Trigonometric Functions. This is unit of Model Textbook of Mathematics for Class XI published by National Book Foundation (NBF) as Federal Textbook Board, Islamabad, Pakistan. 

=====Question 6(i)=====
Find the period: $y=6 \sec(2 x-3)$

** Solution. **

Since period of the $\sec$ is $2\pi$, therefore
\begin{align*}
6 \sec(2 x-3) & = 6 \sec(2 x-3+2\pi) \\
& = 6 \sec(2(x+\pi)-3)
\end{align*}
Hence period of $6 \sec(2 x-3)$ is $\pi$. GOOD
=====Question 6(ii)=====
Find the period: $y=\cos (5 x+4)$

** Solution. **

Since period of the $\cos$ is $2\pi$, therefore
\begin{align*}
\cos (5 x+4) & = 6 \cos(5x+4+2\pi) \\
& = \cos\left(5\left(x+\frac{2\pi}{5}\right)+4\right)
\end{align*}
Hence period of $\cos (5 x+4)$ is $\dfrac{2\pi}{5}$. GOOD
=====Question 6(iii)=====
Find the period: $y=\cot 4 x+\sin \frac{5 x}{2}$

** Solution. **
FIXME(unable to solve)

=====Question 6(iv)=====
Find the period: $y=7 \sin (3 x+3)$

** Solution. **

Since the period of \( \sin \) is \( 2\pi \), therefore:
\begin{align*}
7 \sin(3x + 3) &= 7 \sin(3x + 3 + 2\pi) \\
&= 7 \sin\left(3\left(x + \frac{2\pi}{3}\right) + 3\right).
\end{align*}
Hence, the period of \( 7 \sin(3x + 3) \) is $\dfrac{2\pi}{3}$

=====Question 6(v)=====
Find the period: $y=5 \sin (2 x+3)$

** Solution. **

Since the period of \( \sin \) is \( 2\pi \), therefore:
\begin{align*}
5 \sin(2x + 3) &= 5 \sin(2x + 3 + 2\pi) \\
&= 5 \sin\left(2\left(x + \pi\right) + 3\right).
\end{align*}
Hence, the period of $ 5 \sin(2x + 3)$ is $ \pi.$

=====Question 6(vi)=====
Find the period: $y=2 \tan 3 x+75 \cos x$

** Solution. **

FIXME (problem)

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