Another one vostaomondi747 — let's solve it. 1. Finding the values of $a, b,$ and $c$ Step 1: Substitute the given 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. $$ \frac{P}{Q} = \frac{6 - 2\sqrt{2}}{2} $$ $$ = \frac{6}{2} - \frac{2\sqrt{2}}{2} $$ $$ = 3 - \sqrt{2} $$ Step 5: Compare the result with $a + b\sqrt{c}$. We have $3 - \sqrt{2} = a + b\sqrt{c}$. By comparing the terms, we find: $a = 3$ $b = -1$ $c = 2$ The values are $\boxed{a=3, b=-1, c=2}$. 2. Solving the trigonometric equation Step 1: Use the identity $\sin^2 x = 1 - \cos^2 x$ to rewrite the equation in terms of $\cos x$. $$ 4\sin^2 x + 4\cos x = 5 $$ $$ 4(1 - \cos^2 x) + 4\cos x = 5 $$ Step 2: Expand and rearrange the equation into a quadratic form. $$ 4 - 4\cos^2 x + 4\cos x = 5 $$ $$ -4\cos^2 x + 4\cos x - 1 = 0 $$ Multiply by $-1$ to make the leading coefficient positive: $$ 4\cos^2 x - 4\cos x + 1 = 0 $$ Step 3: Solve the quadratic equation for $\cos x$. This is a perfect square trinomial. $$ (2\cos x - 1)^2 = 0 $$ Take the square root of both sides: $$ 2\cos x - 1 = 0 $$ $$ 2\cos x = 1 $$ $$ \cos x = \frac{1}{2} $$ Step 4: Find the values of $x$ in the range $0^\circ \leq x \leq 360^\circ$. The principal value for $\cos x = \frac{1}{2}$ is $x = 60^\circ$. Since cosine is positive in the first and fourth quadrants, the other solution in the given range is: $$ x = 360^\circ - 60^\circ = 300^\circ $$ The solutions are $\boxed{x = 60^\circ, 300^\circ}$. 3. Making $a$ the subject of the formula Step 1: The given formula is $R = k\sqrt{y^2 - a^2}$. First, isolate the square root term by dividing by $k$. $$ \frac{R}{k} = \sqrt{y^2 - a^2} $$ Step 2: Square both sides of the equation to eliminate the square root. $$ \left(\frac{R}{k}\right)^2 = y^2 - a^2 $$ $$ \frac{R^2}{k^2} = y^2 - a^2 $$ Step 3: Rearrange the equation to isolate $a^2$. Add $a^2$ to both sides and subtract $\frac{R^2}{k^2}$ from both sides. $$ a^2 = y^2 - \frac{R^2}{k^2} $$ Step 4: Take the square root of both sides to find $a$. $$ a = \pm\sqrt{y^2 - \frac{R^2}{k^2}} $$ The subject of the formula is $\boxed{a = \pm\sqrt{y^2 - \frac{R^2}{k^2}}}$. Send me the next one 📸