1. Solve the following trig equations for $0 \leq \theta \leq 360^\circ$. **i)** $2 + 4 \cos^2 \theta = 7 \cos \theta \sin \theta$ Step 1: Use identity $\cos \theta \sin \theta = \frac{1}{2} \sin 2\theta$. Rewrite as: $$2 + 4 \cos^2 \theta = \frac{7}{2} \sin 2\theta$$ Step 2: Express $\cos^2 \theta$ using $\cos 2\theta$: $\cos^2 \theta = \frac{1+\cos 2\theta}{2}$. So, $$2 + 4 \times \frac{1+\cos 2\theta}{2} = \frac{7}{2} \sin 2\theta$$ $$2 + 2 + 2 \cos 2\theta = \frac{7}{2} \sin 2\theta$$ $$4 + 2 \cos 2\theta = \frac{7}{2} \sin 2\theta$$ Step 3: Multiply both sides by 2: $$8 + 4 \cos 2\theta = 7 \sin 2\theta$$ Step 4: Rearrange: $$7 \sin 2\theta - 4 \cos 2\theta = 8$$ Step 5: Let $x = 2\theta$. Then $7 \sin x - 4 \cos x = 8$. Step 6: Write left side as single sine function: $$R \sin(x - \alpha) = 8$$ where $$R = \sqrt{7^2 + (-4)^2} = \sqrt{49 + 16} = \sqrt{65}$$ $$\alpha = \arctan\left(\frac{4}{7}\right)$$ (since sign inside sine is $-4$, so angle is positive to adjust accordingly) Step 7: Then $$\sin(x - \alpha) = \frac{8}{\sqrt{65}} \approx 0.993$$ Since $\sin$ max is 1, $0.993 < 1$, so solutions exist. Step 8: Solve $$x - \alpha = \sin^{-1}(0.993) \quad \text{or} \quad x - \alpha = \pi - \sin^{-1}(0.993)$$ Calculate $\alpha = \arctan(4/7) \approx 29.74^\circ$ $\sin^{-1}(0.993) \approx 83.9^\circ$ So, $$x_1 = 83.9^\circ + 29.74^\circ = 113.64^\circ$$ $$x_2 = 180^\circ - 83.9^\circ + 29.74^\circ = 125.84^\circ$$ Step 9: Since $x = 2\theta$, divide by 2: $$\theta_1 = 56.82^\circ, \quad \theta_2 = 62.92^\circ$$ Also, since sine is periodic every $360^\circ$ in $x$, add $360^\circ$ to $x$ and repeat for $\theta$. Additional solutions in $0 \leq \theta \leq 360^\circ$ are: $$\theta_3 = \frac{113.64^\circ + 360^\circ}{2} = 236.82^\circ$$ $$\theta_4 = \frac{125.84^\circ + 360^\circ}{2} = 242.92^\circ$$ **Solutions:** $\theta = 56.8^\circ, 62.9^\circ, 236.8^\circ, 242.9^\circ$ --- **ii)** Solve $\tan x + \frac{\cos x}{1 + \sin x} = \frac{4}{3}$ Step 1: Simplify $\frac{\cos x}{1 + \sin x}$ by multiplying numerator and denominator by $1 - \sin x$: $$\frac{\cos x (1 - \sin x)}{(1 + \sin x)(1 - \sin x)} = \frac{\cos x (1 - \sin x)}{1 - \sin^2 x}$$ Recall: $1 - \sin^2 x = \cos^2 x$, so $$ = \frac{\cos x (1 - \sin x)}{\cos^2 x} = \frac{1 - \sin x}{\cos x}$$ Step 2: Left side is: $$\tan x + \frac{1 - \sin x}{\cos x} = \frac{\sin x}{\cos x} + \frac{1 - \sin x}{\cos x} = \frac{\sin x + 1 - \sin x}{\cos x} = \frac{1}{\cos x} = \sec x$$ Step 3: So the equation reduces to: $$\sec x = \frac{4}{3} \implies \cos x = \frac{3}{4}$$ Step 4: Solutions for $\cos x = 0.75$ in $0 \leq x \leq 360^\circ$: $$x = \pm \arccos 0.75 + 360^\circ k$$ In degrees: $$x_1 = \arccos 0.75 \approx 41.41^\circ$$ $$x_2 = 360^\circ - 41.41^\circ = 318.59^\circ$$ --- **iii)** Solve $(19 + 2 \sin^2 20^\circ) \tan 20^\circ = \frac{3}{\cos 20^\circ} - 17 \cos 20^\circ$ Step 1: Compute values using degrees: $$\sin 20^\circ \approx 0.3420$$ $$\sin^2 20^\circ \approx 0.116$$ $$19 + 2(0.116) = 19 + 0.232 = 19.232$$ $$\tan 20^\circ \approx 0.3640$$ LHS: $$19.232 \times 0.3640 \approx 7.0$$ Step 2: Calculate RHS: $$\cos 20^\circ \approx 0.9397$$ $$\frac{3}{0.9397} - 17 \times 0.9397 = 3.193 - 15.975 = -12.782$$ Step 3: Since LHS not equal RHS, check problem context or clarify if solving for specific variable or verifying. If solving for $\theta$, this is a numeric evaluation, no variable to solve here. **Interpretation:** This is a verification question; approx values do not equal. --- **iv)** Solve $$\frac{2 + \cos 2x}{3 + \sin^2 2x} = \frac{2}{5}$$ Step 1: Cross-multiply: $$5(2 + \cos 2x) = 2(3 + \sin^2 2x)$$ $$10 + 5 \cos 2x = 6 + 2 \sin^2 2x$$ Step 2: Rearrange: $$5 \cos 2x - 2 \sin^2 2x = 6 - 10 = -4$$ Step 3: Use identity $\sin^2 A = 1 - \cos^2 A$: $$5 \cos 2x - 2(1 - \cos^2 2x) = -4$$ $$5 \cos 2x - 2 + 2 \cos^2 2x = -4$$ Step 4: Rearrange: $$2 \cos^2 2x + 5 \cos 2x - 2 = -4$$ $$2 \cos^2 2x + 5 \cos 2x + 2 = 0$$ Step 5: Let $y = \cos 2x$, quadratic is: $$2 y^2 + 5 y + 2 = 0$$ Step 6: Solve quadratic: $$y = \frac{-5 \pm \sqrt{25 - 16}}{4} = \frac{-5 \pm 3}{4}$$ Step 7: Solutions: $$y_1 = \frac{-5 + 3}{4} = -\frac{1}{2}$$ $$y_2 = \frac{-5 - 3}{4} = -2$$ (invalid as $\cos$ range is [-1,1]) Step 8: So only $\cos 2x = -\frac{1}{2}$ up to $x$ range. Step 9: Solve for $2x$: $$2x = 120^\circ, 240^\circ, 480^\circ (subtract 360^\circ), 600^\circ (subtract 360^\circ)$$ Step 10: Divide by 2: $$x = 60^\circ, 120^\circ, 240^\circ, 300^\circ$$ --- 2. Solve trig equations for $0 \leq \psi \leq 2 \pi$. **i)** $3 \tan \psi + 2 \cos \psi = 0$ Step 1: Write as: $$3 \tan \psi = -2 \cos \psi$$ Step 2: Express tan as $\frac{\sin \psi}{\cos \psi}$: $$3 \frac{\sin \psi}{\cos \psi} = -2 \cos \psi$$ $$3 \sin \psi = -2 \cos^2 \psi$$ Step 3: Use identity $\cos^2 \psi = 1 - \sin^2 \psi$: $$3 \sin \psi = -2 (1 - \sin^2 \psi)$$ $$3 \sin \psi = -2 + 2 \sin^2 \psi$$ Step 4: Rearrange: $$2 \sin^2 \psi - 3 \sin \psi - 2 = 0$$ Step 5: Let $t = \sin \psi$: $$2 t^2 - 3 t - 2 = 0$$ Step 6: Solve quadratic: $$t = \frac{3 \pm \sqrt{9 + 16}}{4} = \frac{3 \pm 5}{4}$$ Step 7: Solutions: $$t_1 = 2, t_2 = -\frac{1}{2}$$ $t_1=2$ invalid since $\sin \psi \in [-1,1]$ Step 8: So $$\sin \psi = -\frac{1}{2}$$ Step 9: Solutions for $\sin \psi = -\frac{1}{2}$ in $[0, 2\pi]$: $$\psi = \frac{7\pi}{6}, \frac{11\pi}{6}$$ **Answers:** $\psi = \frac{7\pi}{6}, \frac{11\pi}{6}$ --- **ii)** Solve $6 \cos \psi = 5 \tan \psi$ Step 1: Write tan as $\frac{\sin \psi}{\cos \psi}$: $$6 \cos \psi = 5 \frac{\sin \psi}{\cos \psi} \implies 6 \cos^2 \psi = 5 \sin \psi$$ Step 2: Use $\cos^2 \psi = 1 - \sin^2 \psi$: $$6 (1 - \sin^2 \psi) = 5 \sin \psi$$ $$6 - 6 \sin^2 \psi = 5 \sin \psi$$ Step 3: Rearrange: $$6 \sin^2 \psi + 5 \sin \psi - 6 = 0$$ Step 4: Let $t = \sin \psi$: $$6 t^2 + 5 t - 6 = 0$$ Step 5: Solve quadratic: $$t = \frac{-5 \pm \sqrt{25 + 144}}{12} = \frac{-5 \pm 13}{12}$$ Step 6: Solutions: $$t_1 = \frac{8}{12} = \frac{2}{3} \approx 0.6667$$ $$t_2 = \frac{-18}{12} = -1.5$$ (invalid) Step 7: So $$\sin \psi = \frac{2}{3}$$ Step 8: Solutions in $[0, 2\pi]$: $$\psi = \arcsin (0.6667) \approx 0.7297 \text{ rad}$$ $$\psi_2 = \pi - 0.7297 = 2.4119 \text{ rad}$$ **Answers:** $\psi \approx 0.73$ rad, $2.41$ rad (to two decimal places) --- **iii)** Solve $(\sqrt{5} + 2 \sin 2\psi)(\sqrt{5} + \tan 2\psi) = 0$ for $0 \leq y \leq \pi$ Step 1: Equation zero when either factor zero: 1) $\sqrt{5} + 2 \sin 2\psi = 0$ or 2) $\sqrt{5} + \tan 2\psi = 0$ Step 2: Solve 1): $$2 \sin 2\psi = -\sqrt{5}$$ $$\sin 2\psi = -\frac{\sqrt{5}}{2}$$ But $|\sin x| \leq 1$, and $\frac{\sqrt{5}}{2} \approx 1.118 > 1$, no solution for 1). Step 3: Solve 2): $$\tan 2\psi = -\sqrt{5}\approx -2.236$$ Step 4: General solutions: $$2\psi = \arctan(-\sqrt{5}) + k\pi$$ Step 5: Calculate principal value: $$\arctan(-2.236) \approx -1.15 \text{ rad}$$ Step 6: Since $0 \leq \psi \leq \pi$, $0 \leq 2\psi \leq 2\pi$, find $k=0,1$: For $k=0$: $$2\psi = -1.15 + \pi = 1.99 \text{ rad}$$ $$\psi = 0.995 \pi$$ (approx) For $k=1$: $$2\psi = -1.15 + 2\pi = 5.13 \text{ rad}$$ $$\psi = 2.56 > \pi$$ not in domain. Step 7: Also add $\pi$ to principal solution: $$2\psi_2 = -1.15 + \pi = 1.99$$ Additional $\psi$ from $2\psi = -1.15 + \pi + \pi = 1.99 + \pi = 5.13$$ But already checked domain. **Answer:** $\psi = 0.995$ rad $\approx \frac{\pi}{3}$ (in terms of $\pi$ it's approximately $0.317\pi$) Note: To express in terms of $\pi$, report approximately as $\psi \approx 0.317 \pi$. --- **iv)** Solve $\tan^4 y - \tan^2 y = 6$ for $0 \leq y \leq 2\pi$ Step 1: Let $t = \tan^2 y$: $$t^2 - t - 6 = 0$$ Step 2: Solve quadratic: $$t = \frac{1 \pm \sqrt{1 + 24}}{2} = \frac{1 \pm 5}{2}$$ Step 3: Solutions: $$t_1 = 3, t_2 = -2$$ (reject $t_2$ as $t = \tan^2 y \geq 0$) Step 4: So $$\tan^2 y = 3 \implies \tan y = \pm \sqrt{3}$$ Step 5: Solve $\tan y = \sqrt{3}$: $$y = \frac{\pi}{3}, \frac{4\pi}{3}$$ Solve $\tan y = -\sqrt{3}$: $$y = \frac{2\pi}{3}, \frac{5\pi}{3}$$ **Answers:** $y = \frac{\pi}{3}, \frac{2\pi}{3}, \frac{4\pi}{3}, \frac{5\pi}{3}$ --- 3. Solve in radians $$\tan(30x - 5)^\circ = \tan 7^\circ, \quad 3 \leq x \leq 6$$ Step 1: Since $\tan A = \tan B$, solutions are $$30x - 5 = 7 + n180$$ Step 2: Rearrange for $x$: $$30x = 12 + 180n$$ $$x = \frac{12 + 180n}{30} = 0.4 + 6n$$ Step 3: Find integer $n$ such that $3 \leq x \leq 6$. For $n=0$, $$x=0.4 < 3$$ For $n=1$, $$x=0.4 + 6 = 6.4 > 6$$ For $n= -1$, $$x=0.4 -6 = -5.6 <3$$ No integer $n$ satisfies, check $n=0$ to $n=1$ for possible non-integer $n$. Step 4: Since the period for tan is $180^\circ$, and coefficient is 30, period in $x$ is $$\Delta x = \frac{180}{30} = 6$$ Try $x=3$ to 6, solutions are at $$x = 0.4 + 6n$$ Closest in domain is at $x=0.4 + 6 = 6.4$ (too big), so $x=0.4 + 0= 0.4$ (too small). No solutions exactly in domain. Step 5: Consider that tangent equation periodic, add multiples to get solutions in domain. Step 6: Correct formula for solutions is: $$30x - 5 = 7 + 180 n$$ For integer $n$. Rewrite: $$30x = 12 + 180 n$$ $$x = \frac{12 + 180 n}{30} = 0.4 + 6n$$ For $x \in [3, 6]$, possible values of $n$ are 0 or 1. At $n=0$, $x=0.4$ (no) At $n=1$, $x=6.4$ (no) Try $n=-1$: $x = 0.4 - 6 = -5.6$ (no) No solutions in domain. Step 7: Thus no solutions satisfy in given domain at exact formula. Confirm by checking near boundaries. **Answer:** No solution in $3 \leq x \leq 6$. --- 4. **i)** Solve $\sin (2\theta + 30^\circ) = \frac{\sqrt{3}}{2}$ for $-180^\circ \leq \theta \leq 180^\circ$ Step 1: Recall $\sin A = \frac{\sqrt{3}}{2}$ at $A = 60^\circ, 120^\circ$ (and periodic every $360^\circ$) Step 2: Set $$2\theta + 30^\circ = 60^\circ + 360^\circ n$$ and $$2\theta + 30^\circ = 120^\circ + 360^\circ n$$ Step 3: Solve for $\theta$: $$\theta = \frac{60^\circ - 30^\circ + 360^\circ n}{2} = 15^\circ + 180^\circ n$$ $$\theta = \frac{120^\circ - 30^\circ + 360^\circ n}{2} = 45^\circ + 180^\circ n$$ Step 4: Find $n$ such that $\theta$ in $[-180^\circ, 180^\circ]$: For $\theta = 15^\circ + 180^\circ n$: - $n=-2$: $15 - 360 = -345^\circ$ no - $n=-1$: $15 - 180 = -165^\circ$ yes - $n=0$: $15^\circ$ yes - $n=1$: $195^\circ$ no For $\theta = 45^\circ + 180^\circ n$: - $n=-2$: $45 - 360 = -315^\circ$ no - $n=-1$: $45 - 180 = -135^\circ$ yes - $n=0$: $45^\circ$ yes - $n=1$: $225^\circ$ no **Solutions:** $\theta = -165^\circ, 15^\circ, -135^\circ, 45^\circ$ --- **ii)** Solve $\sin x = 2 \cos x$ for $0 \leq x \leq 360^\circ$ Step 1: Divide both sides by $\cos x$ (where $\cos x \neq 0$): $$\tan x = 2$$ Step 2: $x = \arctan 2$ and $x = \arctan 2 + 180^\circ$ Calculate: $$x_1 = 63.43^\circ$$ $$x_2 = 243.43^\circ$$ Step 3: Check where $\cos x = 0$: $90^\circ, 270^\circ$ excluded since $\tan$ undefined. **Answers:** $x = 63.4^\circ, 243.4^\circ$ --- **iii)** Solve $$2 \sin^2 y - 5 \cos y + 1 = 0\quad 0 \leq y \leq 2\pi$$ Step 1: Use $\sin^2 y = 1 - \cos^2 y$: $$2(1 - \cos^2 y) - 5 \cos y + 1 = 0$$ $$2 - 2 \cos^2 y - 5 \cos y + 1 = 0$$ $$-2 \cos^2 y - 5 \cos y + 3 = 0$$ Multiply both sides by -1: $$2 \cos^2 y + 5 \cos y - 3 = 0$$ Step 2: Let $t = \cos y$: $$2 t^2 + 5 t - 3 = 0$$ Step 3: Solve quadratic: $$t = \frac{-5 \pm \sqrt{25 + 24}}{4} = \frac{-5 \pm 7}{4}$$ Step 4: Solutions: $$t_1 = \frac{2}{4} = 0.5$$ $$t_2 = \frac{-12}{4} = -3$$ (reject as outside range) Step 5: So $\cos y = 0.5$. Step 6: Solutions for $\cos y = 0.5$ in $0 \leq y \leq 2\pi$: $$y = \frac{\pi}{3}, \quad y = 2\pi - \frac{\pi}{3} = \frac{5\pi}{3}$$ **Answers:** $y = \frac{\pi}{3}, \frac{5\pi}{3}$ --- **Summary:** - Problem 1: 4 solutions for (i), 2 solutions for (ii), no solution (iii), 4 solutions for (iv) - Problem 2: Solutions as found, expressed in radians or $\pi$ as requested - Problem 3: No solution in domain - Problem 4: Solutions as above