answer:We start with the given equation: [ cos(9x) - cos(5x) - sqrt{2} cos(4x) + sin(9x) + sin(5x) = 0. ] 1. Apply trigonometric identities for sum-to-product to simplify the expression. Simplifying the expression: [ cos(9x) - cos(5x) = -2 sinleft(frac{9x + 5x}{2}right) sinleft(frac{9x - 5x}{2}right) = -2 sin(7x) sin(2x), ] [ sin(9x) + sin(5x) = 2 sinleft(frac{9x + 5x}{2}right) cosleft(frac{9x - 5x}{2}right) = 2 sin(7x) cos(2x). ] 2. Substitute these results back into the original equation: [ -2 sin(7x) sin(2x) + 2 sin(7x) cos(2x) - sqrt{2} cos(4x) = 0. ] 3. Factor out (2 sin(7x)) from the equation: [ 2 sin(7x) (cos(2x) - sin(2x)) = sqrt{2} cos(4x). ] 4. Notice that (cos(4x) = cos^2(2x) - sin^2(2x)): [ 2 sin(7x) (cos(2x) - sin(2x)) = sqrt{2} (cos^2(2x) - sin^2(2x)). ] Possible solutions: 5. Case 1: ( sin(7x) = 0 ) [ sin(7x) = 0 Rightarrow 7x = npi Rightarrow x = frac{n pi}{7}, quad n in mathbb{Z}. ] 6. Case 2: ( cos(2x) - sin(2x) = frac{sqrt{2}}{2} (cos(2x) + sin(2x)) ) Use the tangent function to simplify: [ frac{cos(2x) - sin(2x)}{cos(2x) + sin(2x)} = frac{sqrt{2}}{2}. ] Solving for tangent: [ tan(2x - pi/4) = 1 Rightarrow 2x - pi/4 = frac{pi}{4} + kpi Rightarrow 2x = frac{pi}{2} + kpi Rightarrow x = frac{pi}{4} (1 + 2k). ] Hence, [ x = frac{pi}{4} (2k+1), quad k in mathbb{Z}. ] Combining the results: Finally, our combined results include: [ x = frac{n pi}{7}, quad x = frac{pi}{8} + frac{pi k}{2}, quad x = frac{pi}{20} + frac{2pi k}{5}, quad x = frac{pi}{12} + frac{2pi k}{9}, quad k in mathbb{Z}. ] Therefore, the solutions are: [ boxed{x = frac{pi}{8} + frac{pi k}{2}, quad x = frac{pi}{20} + frac{2 pi k}{5}, quad x = frac{pi}{12} + frac{2 pi k}{9}, quad k in mathbb{Z}.} ]

answer:Since a^2 - a < 0, it follows that 0 < a < 1. Also, -a^2 - (-a) = -(a^2 - a) > 0, hence -a^2 > -a. Therefore, we have -a < -a^2 < 0 < a^2 < a. Thus, the answer is boxed{-a < -a^2 < a^2 < a}. This can be obtained by comparing the differences. This problem tests the basic skills of comparing the sizes of inequalities.

answer:Given (mathbf{X}^T = mathbf{X}^{-1}), by definition of matrix inverse, we should have the product (mathbf{X}^T mathbf{X} = mathbf{I}), where (mathbf{I}) is the 3x3 identity matrix. Given: [ mathbf{X}^T = begin{pmatrix} p & s & v q & t & w r & u & x end{pmatrix} ] Multiplying (mathbf{X}^T) with (mathbf{X}), we get: [ begin{pmatrix} p & s & v q & t & w r & u & x end{pmatrix} begin{pmatrix} p & q & r s & t & u v & w & x end{pmatrix} = begin{pmatrix} 1 & 0 & 0 0 & 1 & 0 0 & 0 & 1 end{pmatrix} ] From this matrix product, the diagonal elements result in: - (p^2 + s^2 + v^2 = 1) - (q^2 + t^2 + w^2 = 1) - (r^2 + u^2 + x^2 = 1) Adding these equations, we find: [ p^2 + s^2 + v^2 + q^2 + t^2 + w^2 + r^2 + u^2 + x^2 = 1 + 1 + 1 = boxed{3}. ] Conclusion: Thus, the sum of the squares of all the entries in the matrix (mathbf{X}) is (3).

answer:To find the value of k in the quadratic equation 7x^2+3x+k, given that its roots are frac{-3pm isqrt{299}}{14}, we start by applying the quadratic formula. The quadratic formula gives the roots of a quadratic equation ax^2+bx+c=0 as frac{-bpmsqrt{b^2-4ac}}{2a}. For our equation, a=7, b=3, and c=k, so the roots are: [ frac{-3pmsqrt{3^2-4(7)(k)}}{14} = frac{-3pmsqrt{9-28k}}{14} ] Given that the roots are frac{-3pm isqrt{299}}{14}, we equate the imaginary parts under the square root to find the value of k. This gives us: [ sqrt{9-28k} = isqrt{299} ] Squaring both sides to eliminate the square root and the imaginary unit i, we get: [ 9-28k = -299 ] Solving for k, we have: [ begin{align*} 9 - 28k &= -299 -28k &= -299 - 9 -28k &= -308 k &= frac{-308}{-28} k &= 11 end{align*} ] Therefore, the value of k is boxed{11}.