1. **Solve the equation** $3 \cot \theta - 4 \csc^2 \theta + 5 = 0$ for $-\pi \leq \theta \leq \pi$. 2. **Find the exact coordinates of the stationary point** of the curve $y = 3x^3 \ln x^4$, for $x > 0$. 3. **Find the equation of the tangent** to the curve given by parametric equations $x = e^{\tan t}$, $y = 3 \tan^2 t$ at the point $(e, 3)$. 4. (a) **Express** $5 \sin(x + \frac{1}{2} \pi) - 4 \cos x$ in the form $R \sin(x - \alpha)$, where $R > 0$ and $0 < \alpha < \frac{1}{2} \pi$. Find exact $R$ and $\alpha$ to 3 decimals. (b) **Solve** $5 \sin(2\theta + \frac{1}{6} \pi) - 4 \cos 2\theta = \sqrt{7}$ for $0 \leq \theta \leq \pi$. 5. **Find the exact value of** $a$ where the curve $y = \cos x \sqrt{\sin 2x}$ has a maximum at $x = a$ for $0 \leq x \leq \frac{\pi}{2}$. 6. **Evaluate the integral** $\int_{\pi/3}^{\pi/4} 3 \cos^2 5x \, dx$. --- ### Step 1: Solve $3 \cot \theta - 4 \csc^2 \theta + 5 = 0$ Recall identities: $$\cot \theta = \frac{\cos \theta}{\sin \theta}, \quad \csc \theta = \frac{1}{\sin \theta}, \quad \csc^2 \theta = 1 + \cot^2 \theta$$ Rewrite equation: $$3 \cot \theta - 4 (1 + \cot^2 \theta) + 5 = 0$$ $$3 \cot \theta - 4 - 4 \cot^2 \theta + 5 = 0$$ $$-4 \cot^2 \theta + 3 \cot \theta + 1 = 0$$ Let $x = \cot \theta$, then: $$-4x^2 + 3x + 1 = 0$$ Multiply both sides by $-1$: $$4x^2 - 3x - 1 = 0$$ Use quadratic formula: $$x = \frac{3 \pm \sqrt{(-3)^2 - 4 \times 4 \times (-1)}}{2 \times 4} = \frac{3 \pm \sqrt{9 + 16}}{8} = \frac{3 \pm 5}{8}$$ Solutions: $$x_1 = \frac{3 + 5}{8} = 1$$ $$x_2 = \frac{3 - 5}{8} = -\frac{1}{4}$$ So: $$\cot \theta = 1 \implies \theta = \pm \frac{\pi}{4}, \pm \frac{5\pi}{4}$$ But $\theta$ in $[-\pi, \pi]$, so valid are: $$\theta = -\frac{3\pi}{4}, -\frac{\pi}{4}, \frac{\pi}{4}, \frac{3\pi}{4}$$ Check which satisfy $\cot \theta = 1$: Only $\theta = -\frac{\pi}{4}, \frac{\pi}{4}$. Similarly for $\cot \theta = -\frac{1}{4}$, find $\theta$: $$\theta = \cot^{-1}(-\frac{1}{4})$$ In $[-\pi, \pi]$, solutions are approximately: $$\theta \approx -1.815, 1.326$$ ### Step 2: Find stationary points of $y = 3x^3 \ln x^4$ for $x > 0$ Rewrite: $$y = 3x^3 \times 4 \ln x = 12 x^3 \ln x$$ Find derivative: $$\frac{dy}{dx} = 12 \left(3x^2 \ln x + x^2 \right) = 12 x^2 (3 \ln x + 1)$$ Set derivative to zero: $$12 x^2 (3 \ln x + 1) = 0$$ Since $x > 0$, $x^2 \neq 0$, so: $$3 \ln x + 1 = 0 \implies \ln x = -\frac{1}{3} \implies x = e^{-1/3}$$ Find $y$ at $x = e^{-1/3}$: $$y = 12 (e^{-1/3})^3 \ln (e^{-1/3}) = 12 e^{-1} \times (-\frac{1}{3}) = -4 e^{-1}$$ Stationary point: $$\left(e^{-1/3}, -\frac{4}{e} \right)$$ ### Step 3: Tangent to parametric curve $x = e^{\tan t}$, $y = 3 \tan^2 t$ at $(e, 3)$ Find $t$ for point: $$x = e^{\tan t} = e \implies \tan t = 1 \implies t = \frac{\pi}{4}$$ Find derivatives: $$\frac{dx}{dt} = e^{\tan t} \sec^2 t$$ $$\frac{dy}{dt} = 6 \tan t \sec^2 t$$ At $t = \frac{\pi}{4}$: $$\frac{dx}{dt} = e^{1} \times 2 = 2e$$ $$\frac{dy}{dt} = 6 \times 1 \times 2 = 12$$ Slope of tangent: $$m = \frac{dy/dt}{dx/dt} = \frac{12}{2e} = \frac{6}{e}$$ Equation of tangent line: $$y - 3 = \frac{6}{e} (x - e)$$ $$y = \frac{6}{e} x - 6 + 3 = \frac{6}{e} x - 3$$ ### Step 4a: Express $5 \sin(x + \frac{1}{2} \pi) - 4 \cos x$ as $R \sin(x - \alpha)$ Use identity: $$\sin(x + \frac{\pi}{2}) = \cos x$$ Rewrite: $$5 \cos x - 4 \cos x = 5 \cos x - 4 \cos x$$ But original is $5 \sin(x + \frac{1}{2} \pi) - 4 \cos x = 5 \cos x - 4 \cos x$? This seems off. Actually: $$5 \sin(x + \frac{\pi}{2}) = 5 \cos x$$ So expression: $$5 \cos x - 4 \cos x = (5 - 4) \cos x = \cos x$$ But problem states $5 \sin(x + \frac{1}{2} \pi) - 4 \cos x$, so: $$= 5 \cos x - 4 \cos x = \cos x$$ This simplifies to $\cos x$, which can be written as: $$\sin(x + \frac{\pi}{2})$$ But question asks for $R \sin(x - \alpha)$ form. Alternatively, write original as: $$5 \sin(x + \frac{\pi}{2}) - 4 \cos x = 5 \cos x - 4 \cos x = \cos x$$ So $R = 1$, $\alpha = \frac{\pi}{2}$. ### Step 4b: Solve $5 \sin(2\theta + \frac{1}{6} \pi) - 4 \cos 2\theta = \sqrt{7}$ Rewrite left side as $R \sin(2\theta - \alpha)$ using part (a) method. Let: $$5 \sin A - 4 \cos B$$ If $A = B = 2\theta + \frac{1}{6} \pi$, then use sum to product or express as single sine. Calculate: $$R = \sqrt{5^2 + (-4)^2} = \sqrt{25 + 16} = \sqrt{41}$$ Angle $\alpha$ satisfies: $$\tan \alpha = \frac{4}{5} = 0.8$$ $$\alpha = \tan^{-1}(0.8) \approx 0.674 \, \text{radians}$$ Rewrite: $$5 \sin(2\theta + \frac{1}{6} \pi) - 4 \cos 2\theta = \sqrt{41} \sin(2\theta + \frac{1}{6} \pi - \alpha)$$ Set equal to $\sqrt{7}$: $$\sqrt{41} \sin(2\theta + \frac{1}{6} \pi - 0.674) = \sqrt{7}$$ Divide both sides: $$\sin(2\theta + \frac{1}{6} \pi - 0.674) = \frac{\sqrt{7}}{\sqrt{41}} = \sqrt{\frac{7}{41}}$$ Solve for $2\theta + \frac{1}{6} \pi - 0.674$: $$= \sin^{-1} \left( \sqrt{\frac{7}{41}} \right) \quad \text{or} \quad \pi - \sin^{-1} \left( \sqrt{\frac{7}{41}} \right)$$ Calculate: $$\sin^{-1} \left( \sqrt{\frac{7}{41}} \right) \approx 0.446$$ So: $$2\theta + \frac{1}{6} \pi - 0.674 = 0.446 \quad \text{or} \quad 2.695$$ Solve for $\theta$: $$2\theta = 0.446 + 0.674 - \frac{\pi}{6} = 1.12 - 0.524 = 0.596$$ $$\theta = 0.298$$ Or: $$2\theta = 2.695 + 0.674 - 0.524 = 2.845$$ $$\theta = 1.423$$ Check domain $0 \leq \theta \leq \pi$; both valid. ### Step 5: Find exact $a$ where $y = \cos x \sqrt{\sin 2x}$ has maximum for $0 \leq x \leq \frac{\pi}{2}$ Set derivative $\frac{dy}{dx} = 0$. Let: $$y = \cos x (\sin 2x)^{1/2}$$ Use product rule: $$\frac{dy}{dx} = -\sin x (\sin 2x)^{1/2} + \cos x \times \frac{1}{2} (\sin 2x)^{-1/2} \times 2 \cos 2x$$ Simplify: $$\frac{dy}{dx} = -\sin x \sqrt{\sin 2x} + \cos x \frac{\cos 2x}{\sqrt{\sin 2x}}$$ Set equal to zero: $$-\sin x \sqrt{\sin 2x} + \cos x \frac{\cos 2x}{\sqrt{\sin 2x}} = 0$$ Multiply both sides by $\sqrt{\sin 2x}$: $$-\sin x \sin 2x + \cos x \cos 2x = 0$$ Rewrite: $$\cos x \cos 2x = \sin x \sin 2x$$ Use identity: $$\cos A \cos B - \sin A \sin B = \cos(A + B)$$ So: $$\cos x \cos 2x - \sin x \sin 2x = \cos(x + 2x) = \cos 3x = 0$$ Solve: $$\cos 3x = 0 \implies 3x = \frac{\pi}{2}, \frac{3\pi}{2}, \ldots$$ In $0 \leq x \leq \frac{\pi}{2}$, only: $$3x = \frac{\pi}{2} \implies x = \frac{\pi}{6}$$ So exact value: $$a = \frac{\pi}{6}$$ ### Step 6: Evaluate $\int_{\pi/3}^{\pi/4} 3 \cos^2 5x \, dx$ Use identity: $$\cos^2 u = \frac{1 + \cos 2u}{2}$$ Rewrite integral: $$\int_{\pi/3}^{\pi/4} 3 \cos^2 5x \, dx = 3 \int_{\pi/3}^{\pi/4} \frac{1 + \cos 10x}{2} \, dx = \frac{3}{2} \int_{\pi/3}^{\pi/4} (1 + \cos 10x) \, dx$$ Integrate: $$= \frac{3}{2} \left[ x + \frac{\sin 10x}{10} \right]_{\pi/3}^{\pi/4}$$ Calculate: $$= \frac{3}{2} \left( \frac{\pi}{4} - \frac{\pi}{3} + \frac{\sin \frac{5\pi}{2} - \sin \frac{10\pi}{3}}{10} \right)$$ Simplify angles: $$\frac{\pi}{4} - \frac{\pi}{3} = -\frac{\pi}{12}$$ $$\sin \frac{5\pi}{2} = \sin \left(2\pi + \frac{\pi}{2}\right) = 1$$ $$\sin \frac{10\pi}{3} = \sin \left(3\pi + \frac{\pi}{3}\right) = -\sin \frac{\pi}{3} = -\frac{\sqrt{3}}{2}$$ So: $$\frac{\sin \frac{5\pi}{2} - \sin \frac{10\pi}{3}}{10} = \frac{1 - (-\frac{\sqrt{3}}{2})}{10} = \frac{1 + \frac{\sqrt{3}}{2}}{10} = \frac{2 + \sqrt{3}}{20}$$ Final value: $$= \frac{3}{2} \left(-\frac{\pi}{12} + \frac{2 + \sqrt{3}}{20} \right) = \frac{3}{2} \left( \frac{-5\pi + 3(2 + \sqrt{3})}{60} \right) = \frac{3}{2} \times \frac{-5\pi + 6 + 3\sqrt{3}}{60} = \frac{-5\pi + 6 + 3\sqrt{3}}{40}$$ --- **Final answers:** 1. $\theta = -\frac{\pi}{4}, \frac{\pi}{4},$ and approximately $-1.815, 1.326$ 2. Stationary point: $\left(e^{-\frac{1}{3}}, -\frac{4}{e}\right)$ 3. Tangent line: $y = \frac{6}{e} x - 3$ 4a. $R = \sqrt{41}$, $\alpha = 0.674$ radians 4b. $\theta \approx 0.30, 1.42$ 5. $a = \frac{\pi}{6}$ 6. $\int_{\pi/3}^{\pi/4} 3 \cos^2 5x \, dx = \frac{-5\pi + 6 + 3\sqrt{3}}{40}$