b) \( 3\left(\cos ^{4} x+\sin ^{4} x\right)-2\left(\cos ^{6} x+\sin ^{6} x\right)=1 \)

To solve the equation \( 3(\cos^4 x + \sin^4 x) - 2(\cos^6 x + \sin^6 x) = 1 \), we can utilize trigonometric identities and algebraic manipulation. First, we recognize the identities involving powers of sine and cosine. We start by expressing \( \cos^4 x + \sin^4 x \) and \( \cos^6 x + \sin^6 x \) using the formulas for sums of squares: 1. **Using the identity**: \[ \cos^4 x + \sin^4 x = (\cos^2 x + \sin^2 x)^2 - 2\cos^2 x \sin^2 x = 1 - 2\cos^2 x \sin^2 x \] Therefore, \[ 3(\cos^4 x + \sin^4 x) = 3(1 - 2\cos^2 x \sin^2 x) = 3 - 6\cos^2 x \sin^2 x \] 2. **For the sixth power term**, we can use a similar approach: \[ \cos^6 x + \sin^6 x = (\cos^2 x + \sin^2 x)(\cos^4 x + \sin^4 x) - \cos^2 x \sin^2 x(\cos^2 x + \sin^2 x) = 1(1 - 2\cos^2 x \sin^2 x) - \cos^2 x \sin^2 x \] Thus, \[ \cos^6 x + \sin^6 x = 1 - 3\cos^2 x \sin^2 x \] Accordingly, \[ 2(\cos^6 x + \sin^6 x) = 2(1 - 3\cos^2 x \sin^2 x) = 2 - 6\cos^2 x \sin^2 x \] 3. **Substituting these into the original equation**, we get: \[ 3 - 6\cos^2 x \sin^2 x - (2 - 6\cos^2 x \sin^2 x) = 1 \] Simplifying this gives: \[ 3 - 2 - 6\cos^2 x \sin^2 x + 6\cos^2 x \sin^2 x = 1 \] Which simplifies to: \[ 1 = 1, \] showing that the equation holds for any value of \( x \). Consequently, the equation \( 3(\cos^4 x + \sin^4 x) - 2(\cos^6 x + \sin^6 x) = 1 \) is an identity and is true for all real numbers \( x \).