Prove that $3\cos^2 x + \sin^2 x = 2 + \cos 2x$. Hence, using your result from part (a), Hence, using your result from part (a), find the exact value of $3\cos^2 22.5^\circ + \sin^2 22.5^\circ$. find the exact value of $3\cos^2 22.5^\circ + \sin^2 22.5^\circ$. solve the equation $6\cos^2 x + 2\sin^2 x = 5$ for $0^\circ \le x \le 360^\circ$.

1. Start with the left-hand side and split the cosine squared term

$\text{LHS} = 3\cos^2 x + \sin^2 x$
$= 2\cos^2 x + \cos^2 x + \sin^2 x$

2. Apply the Pythagorean identity

$= 2\cos^2 x + (\cos^2 x + \sin^2 x)$
$= 2\cos^2 x + 1$

3. Apply the double angle formula for cosine to reach the right-hand side

Since $\cos 2x = 2\cos^2 x - 1$, we have $2\cos^2 x = \cos 2x + 1$.
$= (\cos 2x + 1) + 1$
$= 2 + \cos 2x$
$= \text{RHS}$

1. Substitute $x = 22.5^\circ$ into the identity from part (a)

From (a), $3\cos^2 x + \sin^2 x = 2 + \cos 2x$.
Substituting $x = 22.5^\circ$:
$3\cos^2 22.5^\circ + \sin^2 22.5^\circ = 2 + \cos(2 \times 22.5^\circ)$
$= 2 + \cos 45^\circ$

2. Evaluate the exact value using special angles

Since $\cos 45^\circ = \frac{\sqrt{2}}{2}$ (or $\frac{1}{\sqrt{2}}$),
$2 + \cos 45^\circ = 2 + \frac{\sqrt{2}}{2}$

1. Relate the given equation to the identity from part (a)

$6\cos^2 x + 2\sin^2 x = 5$
$2(3\cos^2 x + \sin^2 x) = 5$

2. Substitute the identity to form an equation in $\cos 2x$

From part (a), $3\cos^2 x + \sin^2 x = 2 + \cos 2x$.
$2(2 + \cos 2x) = 5$
$4 + 2\cos 2x = 5$
$2\cos 2x = 1$
$\cos 2x = 0.5$

3. Determine the basic angle and the range for $2x$

Basic angle $\alpha = \cos^{-1}(0.5) = 60^\circ$.
Since $0^\circ \le x \le 360^\circ$, the range for $2x$ is $0^\circ \le 2x \le 720^\circ$.

4. Solve for $2x$ and then $x$

Since $\cos 2x$ is positive, $2x$ lies in the 1st and 4th quadrants.
$2x = 60^\circ, 360^\circ - 60^\circ, 360^\circ + 60^\circ, 720^\circ - 60^\circ$
$2x = 60^\circ, 300^\circ, 420^\circ, 660^\circ$
$x = 30^\circ, 150^\circ, 210^\circ, 330^\circ$