To solve the equation \(7\cos(2x) = 7\sin^2(x) + 2\), we'll use the Pythagorean identity and double-angle identity.
First, let's rewrite the double-angle identity for cosine:
\[ \cos(2x) = 2\cos^2(x) - 1 \]
So, \(7\cos(2x)\) becomes \(7(2\cos^2(x) - 1)\):
\[ 7(2\cos^2(x) - 1) = 7\sin^2(x) + 2 \]
Expanding and simplifying:
\[ 14\cos^2(x) - 7 = 7\sin^2(x) + 2 \]
Next, we'll use the Pythagorean identity \(\sin^2(x) = 1 - \cos^2(x)\):
\[ 14\cos^2(x) - 7 = 7(1 - \cos^2(x)) + 2 \]
Expanding and simplifying further:
\[ 14\cos^2(x) - 7 = 7 - 7\cos^2(x) + 2 \]
Combine like terms:
\[ 14\cos^2(x) - 7 = 9 - 7\cos^2(x) \]
Move all terms to one side of the equation:
\[ 14\cos^2(x) - 7\cos^2(x) = 9 + 7 \]
\[ 7\cos^2(x) = 16 \]
Now, divide both sides by 7:
\[ \cos^2(x) = \frac{16}{7} \]
Finally, take the square root of both sides:
\[ \cos(x) = \pm \sqrt{\frac{16}{7}} \]
\[ \cos(x) = \pm \frac{4}{\sqrt{7}} \]
\[ \cos(x) = \pm \frac{4\sqrt{7}}{7} \]
Now, solve for \(x\) by taking the inverse cosine of both sides:
\[ x = \cos^{-1}\left(\frac{4\sqrt{7}}{7}\right) \]
\[ x = \cos^{-1}\left(-\frac{4\sqrt{7}}{7}\right) \]
Evaluate these expressions to find the approximate values of \(x\).
