Qurition 4 ivlve for \( x \) where \( 0^{\circ} \leq x \leq 90^{\circ} \), in each of the following equations. Give your answer correct. 1) TWO decimal places. \[ \begin{array}{l} \text { 4.shl } \cos 3 x=5,75 \\ 4 . t^{2} \cdot \operatorname{cosec}\left(x-355^{\circ}\right)=7,814 \quad \text { Classwoik } \\ \text { 4ol3, } \frac{\cos \left(2 x+10^{\circ}\right)}{3}=0,159 \end{array} \] 1 \( A=70^{\circ} \) and \( B=40^{\circ} \), use your calculator to evaluate the following(correct to ONE decimal place). \( \therefore 1 \quad \tan (A+B) \) .2. \( 2 \sin ^{2} A \) \( \therefore 3 \sqrt[3]{2 \sec \frac{3 B}{2}} \) thour the use of the cylculator, showing all your working, determine the value of: \( \frac{.11^{2} 60 \cdot \tan 45^{\circ} \cdot \sin 30^{\circ}}{\cos 30^{\circ}} \)

To solve the equations provided, let's start with each individually: 1) **For the equation \( \cos(3x) = 5.75 \)**: The problem lies in the fact that the cosine function can only take values in the range of \([-1, 1]\). Thus, an equation where the cosine equals 5.75 has no solution within the defined range for \(x\). 2) **For the equation \( 4t^2 \cdot \operatorname{cosec}(x - 355^{\circ}) = 7.814 \)**: Rearranging, we get \( \operatorname{cosec}(x - 355^{\circ}) = \frac{7.814}{4t^2} \). Assuming \(t\) is known, one can find the angle by taking the sine inverse of the reciprocal of the right side. 3) **For the equation \( \frac{\cos(2x + 10^\circ)}{3} = 0.159 \)**: This simplifies to \( \cos(2x + 10^\circ) = 0.477 \). Solving for \(2x + 10^\circ\) gives \( 2x = \cos^{-1}(0.477) - 10^\circ \). Divide the resulting angle by 2 to find \(x\). Next, let's calculate the values for \(A = 70^\circ\) and \(B = 40^\circ\): 1) **For \( \tan(A + B) \)**: - \( A + B = 70^\circ + 40^\circ = 110^\circ \). - Therefore, \( \tan(110^\circ) \approx -2.747 \). 2) **Calculate \( 2 \sin^2(A) \)**: - \( \sin(70^\circ) \approx 0.9397 \) - Hence, \( 2(0.9397^2) \approx 1.764 \). 3) **Evaluate \( 3 \sqrt[3]{2 \sec \frac{3B}{2}} \)**: - Calculate \( \frac{3B}{2} = \frac{3 \times 40^\circ}{2} = 60^\circ \). - Thus, \( \sec(60^\circ) = 2 \), leading to \( 3 \sqrt[3]{2 \times 2} = 3 \sqrt[3]{4} \approx 3 \times 1.5874 \approx 4.762 \). Lastly, for the value of \( \frac{.11^{2} \times 60 \times \tan(45^\circ) \times \sin(30^\circ)}{\cos(30^\circ)} \): 1. Since \( \tan(45^\circ) = 1 \) and \( \sin(30^\circ) = 0.5 \), and \( \cos(30^\circ) \approx 0.866 \): 2. The expression simplifies to \( \frac{0.0121 \times 60 \times 1 \times 0.5}{0.866} \approx \frac{0.363}{0.866} \approx 0.419 \). In short, the journey through trigonometric equations can seem daunting, but with a little practice and patience, you can nail these like a boss! Keep calculating and the answers will come running to you! 🚀📐