(i) Find the minimum value of $$ 4+5(2 \sin 3 x-4 \cos 3 x)^{2} $$ (ii) Find the largest value of $$ x $$, for $$ 0 \leqslant x

Question

(i) Find the minimum value of $$ 4+5(2 \sin 3 x-4 \cos 3 x)^{2} $$ (ii) Find the largest value of $$ x $$, for $$ 0 \leqslant x<\pi $$, at which this minimum occurs.

UpStudy AI · Accepted Answer

We start with the expression

$$
4 + 5\bigl(2\sin 3x - 4\cos 3x\bigr)^2.
$$

### (i) Finding the Minimum Value

1. Notice that the expression is in the form

$$
   4 + 5\bigl(\text{something}\bigr)^2.
   $$

Since the square of any real number is nonnegative, we have

$$
   \bigl(2\sin 3x - 4\cos 3x\bigr)^2 \ge 0.
   $$

2. The minimum value of $$\bigl(2\sin 3x - 4\cos 3x\bigr)^2$$ is $$0$$; therefore, the minimum value of the whole expression is

$$
   4 + 5 \cdot 0 = 4.
   $$

### (ii) Finding the Largest $$x$$ in $$0 \le x < \pi$$ at Which This Minimum Occurs

The minimum occurs when

$$
2\sin 3x - 4\cos 3x = 0.
$$

3. Divide both sides by 2 (assuming nonzero)

$$
   \sin 3x - 2\cos 3x = 0.
   $$

4. Rearranging the equation, we get

$$
   \sin 3x = 2\cos 3x.
   $$

5. Dividing by $$\cos 3x$$ (provided $$\cos 3x \neq 0$$):

$$
   \tan 3x = 2.
   $$

6. The general solution for $$3x$$ is

$$
   3x = \arctan 2 + k\pi \quad \text{for any integer } k.
   $$

7. Solving for $$x$$ gives

$$
   x = \frac{\arctan 2 + k\pi}{3}.
   $$

8. Since $$x$$ must lie in the interval $$0 \le x < \pi$$, note that

$$
   3x \text{ ranges over } [0, 3\pi).
   $$

We determine the valid values of $$k$$.

- For $$k = 0$$:

$$
     x = \frac{\arctan 2}{3}.
     $$

- For $$k = 1$$:

$$
     x = \frac{\arctan 2 + \pi}{3}.
     $$

- For $$k = 2$$:

$$
     x = \frac{\arctan 2 + 2\pi}{3}.
     $$

For $$k = 3$$,

$$
     x = \frac{\arctan 2 + 3\pi}{3},
     $$

but this value exceeds $$\pi$$ because

$$
     \frac{3\pi}{3} = \pi,
     $$

and $$\arctan 2 > 0$$.

9. We are asked for the largest $$x$$ in $$0 \le x < \pi$$ for which the minimum occurs. Clearly, the largest occurs for $$k = 2$$:

$$
   x = \frac{\arctan 2 + 2\pi}{3}.
   $$

### Final Answers

(i) The minimum value is $$4$$.

(ii) The largest value of $$x$$ in $$0 \le x < \pi$$ at which this minimum occurs is

$$
x = \frac{\arctan 2 + 2\pi}{3}.
$$
The minimum value is 4, and it occurs at $$ x = \frac{\arctan 2 + 2\pi}{3} $$.

(i) Find the minimum value of

(ii) Find the largest value of ,
for , at which this
minimum occurs.

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