Solve 5/3 − 2x ≤ 1 − 2/3x ≤ 2 − x hence list the integral values that satisfies the inequality

Multiply by 3 to clear the denominators:

Add $6x$ to both sides and subtract 3 from both sides:

Divide by 4:

Multiply by 3 to clear the denominators:

Add $3x$ to both sides and subtract 3 from both sides:

We have $x \geq \frac{1}{2}$ and $x \leq 3$.
So, the solution to the inequality is $\frac{1}{2} \leq x \leq 3$.

The integers between $\frac{1}{2}$ (or 0.5) and 3 (inclusive) are $1, 2, 3$.
The integral values are $\boxed{1, 2, 3}$.

7. Given that $3x$ is an acute angle and $\sin 3x - \cos(x-50^\circ) = 0$, find $\tan(x+35^\circ)$ giving your answer to 4 significant figures.

For $\sin A = \sin B$, the general solutions are $A = B + 360^\circ k$ or $A = 180^\circ - B + 360^\circ k$, where $k$ is an integer.
Case 1: $3x = 140^\circ - x + 360^\circ k$

If $k=0$, $x = 35^\circ$. Then $3x = 3 \times 35^\circ = 105^\circ$. This is not an acute angle ($0^\circ < 3x < 90^\circ$), so this solution is not valid.

Case 2: $3x = 180^\circ - (140^\circ - x) + 360^\circ k$

If $k=0$, $x = 20^\circ$. Then $3x = 3 \times 20^\circ = 60^\circ$. This is an acute angle ($0^\circ < 60^\circ < 90^\circ$), so this is the correct value for $x$.

Substitute $x=20^\circ$:

Using a calculator:

Rounding to 4 significant figures:

The value is $\boxed{1.428}$.

8. Solve for $y$ in the equation $3^{6y} - 4(3^{3y}) - 5 = 40$.

Then $3^{6y} = (3^{3y})^2 = u^2$. Substitute $u$ into the equation to form a quadratic equation.

Find two numbers that multiply to -45 and add to -4. These numbers are -9 and 5.

This gives two possible values for $u$:

Case 1: $3^{3y} = 9$

Equate the exponents:

Case 2: $3^{3y} = -5$
An exponential function with a positive base (like 3) can never produce a negative result. Therefore, there is no real solution for $y$ in this case.

The only real solution for $y$ is $\boxed{\frac{2}{3}}$.