Question 5 5.1 If $$ 5 \sin x=4 $$ and $$ x \in\left[0^{\circ} ; 90^{\circ}\right] $$; use the sketch to calculate the value of the following: 5.1.1 $$ \frac{\sin x}{\cos x} $$ $$ 5.1 .2 \sin ^{2} x+\cos ^{2} x $$ 5.2 Solve for $$ x $$ and $$ x \in\left[0^{\circ} ; 90^{\circ}\right] $$; $$ \sin x=0,3 $$

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**Step 1. Solve for $$\sin x$$ from $$5\sin x = 4$$**

We have:
$$
5\sin x = 4 \quad \Longrightarrow \quad \sin x = \frac{4}{5}
$$

**Step 2. Calculate $$\frac{\sin x}{\cos x}$$**

We know that:
$$
\tan x = \frac{\sin x}{\cos x}
$$
To compute $$\cos x$$, use the Pythagorean identity:
$$
\sin^2 x + \cos^2 x = 1
$$
Substitute $$\sin x = \frac{4}{5}$$:
$$
\left(\frac{4}{5}\right)^2 + \cos^2 x = 1 \quad \Longrightarrow \quad \frac{16}{25} + \cos^2 x = 1
$$
Solve for $$\cos^2 x$$:
$$
\cos^2 x = 1 - \frac{16}{25} = \frac{25}{25} - \frac{16}{25} = \frac{9}{25}
$$
Since $$x \in [0^\circ, 90^\circ]$$, $$\cos x$$ is positive, so:
$$
\cos x = \frac{3}{5}
$$
Now compute:
$$
\frac{\sin x}{\cos x} = \frac{\frac{4}{5}}{\frac{3}{5}} = \frac{4}{3}
$$

**Step 3. Verify $$\sin^2 x + \cos^2 x$$**

Using the Pythagorean identity:
$$
\sin^2 x + \cos^2 x = 1
$$
Thus,
$$
\sin^2 x + \cos^2 x = 1
$$

**Step 4. Solve for $$x$$ given $$\sin x = 0.3$$ and $$x \in [0^\circ, 90^\circ]$$**

Take the inverse sine (arcsin) of both sides:
$$
x = \arcsin(0.3)
$$
Using a calculator (ensuring the mode is in degrees):
$$
x \approx 17.46^\circ
$$

**Final Answers:**

- $$ \frac{\sin x}{\cos x} = \frac{4}{3} $$
- $$ \sin^2 x + \cos^2 x = 1 $$
- $$ x \approx 17.46^\circ $$ when $$\sin x = 0.3$$ in the interval $$[0^\circ, 90^\circ]$$
- $$ \frac{\sin x}{\cos x} = \frac{4}{3} $$ - $$ \sin^2 x + \cos^2 x = 1 $$ - $$ x \approx 17.46^\circ $$

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