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 \( x \in\left[0^{\circ} ; 90^{\circ}\right] \); \( \sin x=0,3 \)

**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 \([0^\circ, 90^\circ]\)