We are tasked with finding the range of $x$ in the interval $[0, 2\pi]$ for which the inequality $3\sin x + 2\cos x > 3$ holds, using the t-formulae. Let's break it down step by step:

Step 1: Rewrite the inequality using the t-formulae

The t-formulae involve the substitution: $t = \tan \left(\frac{x}{2}\right)$ With this substitution, we can express $\sin x$ and $\cos x$ in terms of $t$ as follows: $\sin x = \frac{2t}{1 + t^2}, \quad \cos x = \frac{1 - t^2}{1 + t^2}$ Substitute these into the inequality $3\sin x + 2\cos x > 3$.

Step 2: Apply the substitution to the inequality

Substituting the t-formulae expressions into the inequality:

$3 \cdot \frac{2t}{1 + t^2} + 2 \cdot \frac{1 - t^2}{1 + t^2} > 3$

Simplify the left-hand side: $\frac{6t + 2(1 - t^2)}{1 + t^2} > 3$ This simplifies to: $\frac{6t + 2 - 2t^2}{1 + t^2} > 3$

Step 3: Eliminate the denominator

Multiply both sides by $1 + t^2$ to get rid of the denominator: $6t + 2 - 2t^2 > 3(1 + t^2)$ Expanding both sides: $6t + 2 - 2t^2 > 3 + 3t^2$

Step 4: Simplify the inequality

Bring all terms to one side: $-2t^2 - 3t^2 + 6t + 2 - 3 > 0$ This simplifies to: $-5t^2 + 6t - 1 > 0$

Step 5: Solve the quadratic inequality

We now solve the quadratic inequality $-5t^2 + 6t - 1 > 0$. First, solve the corresponding quadratic equation: $-5t^2 + 6t - 1 = 0$ Using the quadratic formula: $t = \frac{-6 \pm \sqrt{6^2 - 4(-5)(-1)}}{2(-5)} = \frac{-6 \pm \sqrt{36 - 20}}{-10} = \frac{-6 \pm \sqrt{16}}{-10}$ $t = \frac{-6 \pm 4}{-10}$ So, the solutions for $t$ are: $t = \frac{-6 + 4}{-10} = \frac{-2}{-10} = \frac{1}{5}, \quad t = \frac{-6 - 4}{-10} = \frac{-10}{-10} = 1$

Thus, the quadratic equation has roots at $t = \frac{1}{5}$ and $t = 1$.

Step 6: Analyze the sign of the quadratic

The inequality $-5t^2 + 6t - 1 > 0$ is positive between the roots $\frac{1}{5}$ and $1$. Therefore, $t$ must lie in the interval: $\frac{1}{5} < t < 1$

Step 7: Convert back to $x$

Recall that $t = \tan \left( \frac{x}{2} \right)$, so we have: $\frac{1}{5} < \tan \left( \frac{x}{2} \right) < 1$

Now solve for $x$:

1. For $\tan \left( \frac{x}{2} \right) = \frac{1}{5}$: $\frac{x}{2} = \tan^{-1} \left( \frac{1}{5} \right) \quad \Rightarrow \quad x = 2 \tan^{-1} \left( \frac{1}{5} \right)$ Using a calculator, $\tan^{-1} \left( \frac{1}{5} \right) \approx 0.197$. So, $x \approx 2 \times 0.197 = 0.394$.

2. For $\tan \left( \frac{x}{2} \right) = 1$: $\frac{x}{2} = \frac{\pi}{4} \quad \Rightarrow \quad x = 2 \times \frac{\pi}{4} = \frac{\pi}{2}$

Thus, the range of $x$ is approximately: $0.394 < x < \frac{\pi}{2}$

Final Answer:

The range of $x$ for which $3\sin x + 2\cos x > 3$ is: $x \in (0.394, \frac{\pi}{2})$

Would you like further clarification on any of the steps?

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Here are 5 related questions:

1. How do we solve trigonometric inequalities involving more complex expressions?
2. What are the advantages of using the t-formulae in solving trigonometric equations?
3. How can we use graphical methods to solve trigonometric inequalities?
4. What are the general steps to convert between trigonometric expressions and t-formulae?
5. Can we extend this method to other intervals, like $[-\pi, \pi]$?

Tip: When solving quadratic inequalities, always consider the sign of the quadratic expression between its roots.