Here are the solutions to the problems: Problem 1: Given $P = 4 + \sqrt{2}$ and $Q = 2 + \sqrt{2}$. We need to find $a, b, c$ such that $\frac{P}{Q} = a + b\sqrt{c}$. Step 1: Substitute the values of $P$ and $Q$ into the expression $\frac{P}{Q}$. $$ \frac{P}{Q} = \frac{4 + \sqrt{2}}{2 + \sqrt{2}} $$ Step 2: Rationalize the denominator by multiplying the numerator and denominator by the conjugate of the denominator, which is $2 - \sqrt{2}$. $$ \frac{P}{Q} = \frac{4 + \sqrt{2}}{2 + \sqrt{2}} \times \frac{2 - \sqrt{2}}{2 - \sqrt{2}} $$ Step 3: Expand the numerator and the denominator. Numerator: $$ (4 + \sqrt{2})(2 - \sqrt{2}) = 4(2) - 4\sqrt{2} + 2\sqrt{2} - (\sqrt{2})^2 $$ $$ = 8 - 4\sqrt{2} + 2\sqrt{2} - 2 $$ $$ = 6 - 2\sqrt{2} $$ Denominator: $$ (2 + \sqrt{2})(2 - \sqrt{2}) = 2^2 - (\sqrt{2})^2 $$ $$ = 4 - 2 $$ $$ = 2 $$ Step 4: Combine the simplified numerator and denominator. $$ \frac{P}{Q} = \frac{6 - 2\sqrt{2}}{2} $$ Step 5: Simplify the fraction. $$ \frac{P}{Q} = \frac{6}{2} - \frac{2\sqrt{2}}{2} $$ $$ \frac{P}{Q} = 3 - \sqrt{2} $$ Step 6: Compare this result with $a + b\sqrt{c}$. $$ 3 - \sqrt{2} = a + b\sqrt{c} $$ By comparison, we have: $a = 3$ $b = -1$ $c = 2$ The values are $\boxed{a = 3, b = -1, c = 2}$. --- Problem 2: Solve the equation $4\sin^2 x + 4\cos x = 5$ for $0^\circ \le x \le 360^\circ$. Step 1: Use the trigonometric identity $\sin^2 x + \cos^2 x = 1$, which implies $\sin^2 x = 1 - \cos^2 x$. Substitute this into the equation. $$ 4(1 - \cos^2 x) + 4\cos x = 5 $$ Step 2: Expand and rearrange the equation into a standard quadratic form $A\cos^2 x + B\cos x + C = 0$. $$ 4 - 4\cos^2 x + 4\cos x = 5 $$ $$ -4\cos^2 x + 4\cos x + 4 - 5 = 0 $$ $$ -4\cos^2 x + 4\cos x - 1 = 0 $$ Multiply by -1 to make the leading coefficient positive: $$ 4\cos^2 x - 4\cos x + 1 = 0 $$ Step 3: Let $y = \cos x$. The equation becomes a quadratic equation in $y$: $$ 4y^2 - 4y + 1 = 0 $$ This is a perfect square trinomial: $(2y - 1)^2 = 0$. Alternatively, use the quadratic formula: $y = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(4)(1)}}{2(4)}$ $$ y = \frac{4 \pm \sqrt{16 - 16}}{8} $$ $$ y = \frac{4 \pm 0}{8} $$ $$ y = \frac{4}{8} $$ $$ y = \frac{1}{2} $$ Step 4: Substitute back $y = \cos x$. $$ \cos x = \frac{1}{2} $$ Step 5: Find the values of $x$ in the range $0^\circ \le x \le 360^\circ$ for which $\cos x = \frac{1}{2}$. The cosine function is positive in the first and fourth quadrants. The reference angle is $\cos^{-1}\left(\frac{1}{2}\right) = 60^\circ$. In the first quadrant: $$ x = 60^\circ $$ In the fourth quadrant: $$ x = 360^\circ - 60^\circ = 300^\circ $$ The solutions are $\boxed{x = 60^\circ, 300^\circ}$. --- Problem 3: Make $a$ the subject of the formula: $R = h\sqrt{y^2 - a^2}$. Step 1: Divide both sides by $h$ to isolate the square root term. $$ \frac{R}{h} = \sqrt{y^2 - a^2} $$ Step 2: Square both sides of the equation to eliminate the square root. $$ \left(\frac{R}{h}\right)^2 = (\sqrt{y^2 - a^2})^2 $$ $$ \frac{R^2}{h^2} = y^2 - a^2 $$ Step 3: Rearrange the equation to isolate $a^2$. Add $a^2$ to both sides and subtract $\frac{R^2}{h^2}$ from both sides. $$ a^2 = y^2 - \frac{R^2}{h^2} $$ Step 4: Take the square root of both sides to solve for $a$. Remember to include both positive and negative roots. $$ a = \pm\sqrt{y^2 - \frac{R^2}{h^2}} $$ This can also be written with a common denominator inside the square root: $$ a = \pm\sqrt{\frac{y^2h^2 - R^2}{h^2}} $$ $$ a = \pm\frac{\sqrt{y^2h^2 - R^2}}{h} $$ The subject of the formula is $\boxed{a = \pm\sqrt{y^2 - \frac{R^2}{h^2}}}$ or $\boxed{a = \pm\frac{\sqrt{y^2h^2 - R^2}}{h}}$.