1. **Problem statement:** Find $x$ for $0^\circ < x < 360^\circ$ given: (i) $\sin x = -0.3782$ and $\cos x > 0$. (ii) $\tan x = -2.361$ and $\cos x < 0$. (iii) $\cos x = -0.4713$ and $\tan x > 0$. (iv) $\sin \frac{x}{2} = 0.1796$ and $\tan x < 0$. 2. **Recall key trigonometric rules:** - $\sin x$ is negative in Quadrants III and IV. - $\cos x$ is positive in Quadrants I and IV. - $\tan x = \frac{\sin x}{\cos x}$. - $\tan x$ is negative when $\sin x$ and $\cos x$ have opposite signs. - For $\sin \theta = a$, solutions in $0^\circ$ to $360^\circ$ are $\theta = \sin^{-1}(a)$ and $180^\circ - \sin^{-1}(a)$. - For $\cos \theta = a$, solutions are $\theta = \cos^{-1}(a)$ and $360^\circ - \cos^{-1}(a)$. - For $\tan \theta = a$, solutions are $\theta = \tan^{-1}(a)$ and $\theta = 180^\circ + \tan^{-1}(a)$. 3. **Solve each part:** (i) $\sin x = -0.3782$, $\cos x > 0$. - $\sin x$ negative means $x$ in Quadrants III or IV. - $\cos x > 0$ means Quadrants I or IV. - Intersection: Quadrant IV. - Find reference angle: $\alpha = \sin^{-1}(0.3782) \approx 22.2^\circ$. - Since $x$ in Quadrant IV, $x = 360^\circ - 22.2^\circ = 337.8^\circ$. (ii) $\tan x = -2.361$, $\cos x < 0$. - $\tan x$ negative means $x$ in Quadrants II or IV. - $\cos x < 0$ means Quadrants II or III. - Intersection: Quadrant II. - Find reference angle: $\alpha = \tan^{-1}(2.361) \approx 67.5^\circ$. - Since $x$ in Quadrant II, $x = 180^\circ - 67.5^\circ = 112.5^\circ$. (iii) $\cos x = -0.4713$, $\tan x > 0$. - $\cos x$ negative means Quadrants II or III. - $\tan x > 0$ means Quadrants I or III. - Intersection: Quadrant III. - Find reference angle: $\alpha = \cos^{-1}(0.4713) \approx 61.9^\circ$. - Since $x$ in Quadrant III, $x = 180^\circ + 61.9^\circ = 241.9^\circ$. (iv) $\sin \frac{x}{2} = 0.1796$, $\tan x < 0$. - First find $\frac{x}{2}$. - $\sin \theta = 0.1796$ gives $\theta_1 = \sin^{-1}(0.1796) \approx 10.34^\circ$, $\theta_2 = 180^\circ - 10.34^\circ = 169.66^\circ$. - So $\frac{x}{2} = 10.34^\circ$ or $169.66^\circ$. - Thus $x = 20.68^\circ$ or $339.32^\circ$. - Now check $\tan x < 0$. - $\tan 20.68^\circ > 0$ (Quadrant I), discard. - $\tan 339.32^\circ$ is $\tan (360^\circ - 20.68^\circ)$, which is negative (Quadrant IV), keep. 4. **Final answers:** (i) $x = 337.8^\circ$ (ii) $x = 112.5^\circ$ (iii) $x = 241.9^\circ$ (iv) $x = 339.32^\circ$