answer:To solve for the possible values of angle C in degrees, we start by squaring both given equations: 1. Squaring the first equation, 3 sin A + 4 cos B = 6, we get: [9 sin^2 A + 24 sin A cos B + 16 cos^2 B = 36.] 2. Squaring the second equation, 4 sin B + 3 cos A = 1, we obtain: [16 sin^2 B + 24 cos A sin B + 9 cos^2 A = 1.] Adding these two equations together and using the trigonometric identity cos^2 theta + sin^2 theta = 1, we simplify to: [24 sin A cos B + 24 cos A sin B = 12.] Dividing both sides by 24, we find: [sin A cos B + cos A sin B = frac{1}{2}.] Using the angle addition formula, sin (A + B) = sin A cos B + cos A sin B, we deduce: [sin (A + B) = frac{1}{2}.] Since sin C = sin (180^circ - A - B), we have: [sin C = sin (A + B) = frac{1}{2}.] Therefore, the possible values for angle C are 30^circ or 150^circ. However, if angle C = 150^circ, then angle A < 30^circ, which implies: [3 sin A + 4 cos B < 3 cdot frac{1}{2} + 4 < 6,] leading to a contradiction with the initial condition. Thus, the only feasible value for angle C is boxed{30^circ}. This conclusion is supported by the existence of a triangle ABC that satisfies the given conditions, with specific values for cos A and cos B.

answer:1. **Define Variables and Recollect the Problem:** - We are given a right-angled triangle ( triangle ABC ) with ( widehat{ACB} = 90^circ ), hypotenuse ( AB = c ), and a bisector ( AD ) of the acute angle ( widehat{CAB} ) such that ( |AD| = frac{c sqrt{3}}{3} ). - We are required to find the lengths of the legs ( AC ) and ( BC ). 2. **Assign Known Values:** - Let ( |AC| = x ) and ( |BC| = y ). 3. **Apply the Pythagorean Theorem:** - Because ( triangle ABC ) is a right triangle: [ x^2 + y^2 = c^2 ] 4. **Determine Additional Variables and Setup Further Equations:** - Introduce the auxiliary unknown ( |CD| = z ). - Given that ( AD ) is the angle bisector, we have the angle bisector theorem, but it isn't directly applicable here. Instead, use geometric properties and trigonometric identities: [ |AD| = frac{c sqrt{3}}{3} ] 5. **Translate ( AD ) (Bisector) into Known Quantities Using Angle:** - Assume ( widehat{CAD} = widehat{BAD} = t ). Use the property of angle bisectors and trigonometry to form: [ |AC| = x = frac{c sqrt{3}}{3} cos t ] [ |BC| = y = c sin 2t ] 6. **Combine Them into the Pythagorean Theorem:** - Substitute these expressions back into the Pythagorean theorem: [ left( frac{c sqrt{3}}{3} cos t right)^2 + left( c sin 2t right)^2 = c^2 ] 7. **Simplify and Solve for Trigonometric Function ( t ):** - Simplify the above equation: [ frac{c^2}{3} cos^2 t + c^2 sin^2 2t = c^2 ] [ frac{1}{3} cos^2 t + sin^2 2t = 1 ] We know that ( sin 2t = 2 sin t cos t ), so: [ cos^2 t + 12 left( sin t cos t right)^2 = 3 ] [ cos^2 t + 12 sin^2 t cos^2 t = 3 ] 8. **Use the Double Angle Identity:** - Rewrite and solve for ( cos 2t ): [ cos^2 t + 3 left(1 - cos^2 2t right) = 3 ] [ 6 cos^2 2t - cos 2t - 1 = 0 ] 9. **Solve Quadratic for ( cos 2t ):** - Use the quadratic formula ( ax^2 + bx + c = 0 ): [ ( cos 2t )_1 = -frac{1}{3}, quad ( cos 2t )_2 = frac{1}{2} ] 10. **Select Valid Solution:** - Given ( 0^circ < 2t < 90^circ ), ( cos 2t = frac{1}{2} ) is valid: [ cos 2t = frac{1}{2} implies t = 30^circ ] 11. **Find the Sides ( |AC| ) and ( |BC| ):** - Substitute ( t = 30^circ ): [ |AC| = x = frac{c sqrt{3}}{3} cos 30^circ = frac{c sqrt{3}}{3} times frac{sqrt{3}}{2} = frac{c}{2} ] [ |BC| = y = c sin 60^circ = c times frac{sqrt{3}}{2} = frac{c sqrt{3}}{2} ] **Conclusion:** [ |AC| = frac{c}{2} quad text{and} quad |BC| = frac{c sqrt{3}}{2} ] [ boxed{frac{c}{2}, frac{c sqrt{3}}{2}} ]

answer:To find the constant term in the expansion of left(x^3 - frac{1}{x}right)^4, we first identify the general term of the expansion, which is given by: [T_{r+1} = C_4^r left(x^3right)^{4-r} left(-frac{1}{x}right)^r = C_4^r (-1)^r x^{12-4r}] Here, r can take any value from 0 to 4. For the term to be constant, the exponent of x must be 0. Therefore, we set the exponent equal to 0 and solve for r: [12 - 4r = 0] Solving for r gives: [4r = 12 implies r = 3] Substituting r = 3 into the formula for the general term to find the constant term: [T_4 = C_4^3 (-1)^3 = -4] Therefore, the constant term in the expansion is boxed{-4}.

answer:To show that the sequence {x_n} has a finite limit and to determine this limit, we will proceed as follows: 1. **Define the sequence and its properties:** Given the sequence {x_n} with initial values: [ x_1 = 1, quad x_2 = 9, quad x_3 = 9, quad x_4 = 1 ] and the recurrence relation for n geq 1: [ x_{n+4} = sqrt[4]{x_n cdot x_{n+1} cdot x_{n+2} cdot x_{n+3}} ] 2. **Transform the sequence using logarithms:** Define a new sequence {y_n} such that: [ y_n = log_3 x_n ] This transforms the recurrence relation into: [ y_{n+4} = frac{y_n + y_{n+1} + y_{n+2} + y_{n+3}}{4} ] with initial values: [ y_1 = log_3 1 = 0, quad y_2 = log_3 9 = 2, quad y_3 = log_3 9 = 2, quad y_4 = log_3 1 = 0 ] 3. **Analyze the transformed sequence:** The recurrence relation for {y_n} is a linear homogeneous recurrence relation. We can write it in matrix form: [ begin{pmatrix} y_{n+4} y_{n+3} y_{n+2} y_{n+1} end{pmatrix} = begin{pmatrix} frac{1}{4} & frac{1}{4} & frac{1}{4} & frac{1}{4} 1 & 0 & 0 & 0 0 & 1 & 0 & 0 0 & 0 & 1 & 0 end{pmatrix} begin{pmatrix} y_n y_{n+1} y_{n+2} y_{n+3} end{pmatrix} ] 4. **Find the eigenvalues of the matrix:** The characteristic polynomial of the matrix is: [ detleft( begin{pmatrix} frac{1}{4} & frac{1}{4} & frac{1}{4} & frac{1}{4} 1 & 0 & 0 & 0 0 & 1 & 0 & 0 0 & 0 & 1 & 0 end{pmatrix} - lambda I right) = 0 ] Solving this, we find the eigenvalues to be lambda = 1, -i, i, -1. 5. **Analyze the long-term behavior:** The eigenvalue lambda = 1 indicates that the sequence {y_n} will converge to a constant value as n to infty. The other eigenvalues (-i, i, -1) have magnitudes less than or equal to 1, which means their contributions will diminish over time. 6. **Determine the limit:** Since {y_n} converges to a constant value, let L be this limit. Then: [ L = frac{L + L + L + L}{4} implies L = L ] This confirms that {y_n} converges to a constant value. Given the initial values and the symmetry, the limit L must be the average of the initial values: [ L = frac{0 + 2 + 2 + 0}{4} = 1 ] 7. **Transform back to the original sequence:** Since y_n = log_3 x_n and y_n to 1, we have: [ log_3 x_n to 1 implies x_n to 3^1 = 3 ] The final answer is boxed{3}