Question

question 5
solve ( 2cos^{2} u - 7cos u + 5 = 0 ) for all solutions.
example solution input: if ( \frac{3pi}{4} ) and ( pi ) are solutions, input 3pi/4, pi.
answer: ( x = square + 2pi k ), where ( k ) is an (square) (natural number, integer, rational number, real number, or complex number)

• a solution write - up is required for this problem.

Explanation:

Step1: Let \( t = \cos u \)

Substitute \( \cos u \) with \( t \) in the equation \( 2\cos^2 u - 7\cos u + 5 = 0 \), we get the quadratic equation \( 2t^2 - 7t + 5 = 0 \).

Step2: Solve the quadratic equation

Factor the quadratic equation: \( 2t^2 - 7t + 5 = (2t - 5)(t - 1) = 0 \).
Set each factor equal to zero:
- \( 2t - 5 = 0 \) gives \( t = \frac{5}{2} \). But \( \cos u \) has a range of \([-1, 1]\), so \( t = \frac{5}{2} \) is not a valid solution.
- \( t - 1 = 0 \) gives \( t = 1 \).

Step3: Find \( u \) when \( \cos u = 1 \)

The general solution for \( \cos u = 1 \) is \( u = 2k\pi \), where \( k \) is an integer. Wait, wait, no, wait. Wait, let's check the quadratic solution again. Wait, maybe I made a mistake in factoring. Let's use the quadratic formula. For \( ax^2 + bx + c = 0 \), \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). Here, \( a = 2 \), \( b = -7 \), \( c = 5 \). So \( t = \frac{7 \pm \sqrt{49 - 40}}{4} = \frac{7 \pm 3}{4} \). So \( t = \frac{7 + 3}{4} = \frac{10}{4} = \frac{5}{2} \) (invalid) or \( t = \frac{7 - 3}{4} = 1 \) (valid). Wait, but maybe I misread the original equation. Wait, the original equation is \( 2\cos^2 u - 7\cos u + 5 = 0 \)? Wait, the user wrote "2 cos² u − 7 cos u + 5 = 0". Wait, maybe I made a mistake. Wait, let's check again. Wait, maybe the equation is \( 2\cos^2 u - 7\cos u + 3 = 0 \)? No, the user wrote 5. Wait, but if \( t = 1 \), then \( \cos u = 1 \), so \( u = 2k\pi \). But that seems too simple. Wait, maybe I misread the equation. Wait, the user's equation: "2 cos² u − 7 cos u + 5 = 0". Let me check the quadratic again. \( 2t^2 -7t +5 \). Discriminant: \( 49 - 40 = 9 \). So roots are \( (7 \pm 3)/4 \), so 10/4=2.5 (invalid) and 4/4=1 (valid). So only \( \cos u = 1 \), so \( u = 2k\pi \). But that seems odd. Wait, maybe the original equation was \( 2\cos^2 u - 7\cos u + 3 = 0 \)? Let's check. If it's 3, then discriminant is 49 - 24 = 25, roots (7±5)/4, so 12/4=3 (invalid) or 2/4=0.5 (valid). Then \( \cos u = 0.5 \), so \( u = \frac{\pi}{3} + 2k\pi \) or \( u = \frac{5\pi}{3} + 2k\pi \). But the user's equation is 5. Wait, maybe the user made a typo, but according to the given equation, the only valid solution is \( \cos u = 1 \), so \( u = 2k\pi \). But let's check again. Wait, the original problem: "Solve \( 2\cos^2 u - 7\cos u + 5 = 0 \) for all solutions." So with \( t = \cos u \), quadratic in t: \( 2t^2 -7t +5 = 0 \). Solutions t=1 and t=5/2 (invalid). So \( \cos u = 1 \), so \( u = 2k\pi \), where k is an integer. But that seems too simple. Wait, maybe I misread the equation. Wait, maybe it's \( 2\cos^2 u - 7\cos u - 5 = 0 \)? Then discriminant is 49 + 40 = 89, roots (7±√89)/4, which are not in [-1,1]. No. Wait, maybe the equation is \( 2\cos^2 u + 7\cos u + 5 = 0 \)? Then discriminant 49 - 40 = 9, roots (-7±3)/4, so (-4)/4=-1 or (-10)/4=-2.5 (invalid). So \( \cos u = -1 \), so \( u = \pi + 2k\pi \). Ah! Wait, maybe the original equation is \( 2\cos^2 u + 7\cos u + 5 = 0 \)? But the user wrote "−7 cos u". Wait, the user's equation: "2 cos² u − 7 cos u + 5 = 0". So it's minus 7 cos u. So my initial solution is correct. So \( \cos u = 1 \), so \( u = 2k\pi \). But that seems odd. Maybe the user made a typo, but according to the given equation, that's the solution.

Wait, but the example solution input is \( 3\pi/4, \pi \), so maybe the original equation was different. Wait, maybe the equation is \( 2\cos^2 u - 7\cos u + 3 = 0 \)? Let's assume that for a moment. If it's 3, then \( 2t^2 -7t +3 = 0 \), discriminant 49 - 24 = 25, roots (7±5)/4, so 12/4=3 (invalid) or 2/4=0.5 (valid). Then \( \cos u = 0.5 \), so \( u = \frac{\pi}{3} + 2k\pi \) or \( u = \frac{5\pi}{3} + 2k\pi \). But the user's equation is 5. Alternatively, maybe the equation is \( 2\cos^2 u - 5\cos u + 3 = 0 \)? Then discriminant 25 - 24 = 1, roots (5±1)/4, so 6/4=1.5 (invalid) or 4/4=1 (valid). No. Wait, maybe the original equation is \( 2\cos^2 u - 7\cos u + 5 = 0 \), and the solution is \( u = 2k\pi \). But that seems too simple. Alternatively, maybe I made a mistake in the sign. Wait, the equation is \( 2\cos^2 u - 7\cos u + 5 = 0 \). Let's plug \( u = \pi \): \( 2(1) -7(-1) +5 = 2 +7 +5 = 14 eq 0 \). Plug \( u = 0 \): \( 2(1) -7(1) +5 = 0 \). Yes! \( u = 0 \) is a solution, and \( 0 = 2k\pi \) when k=0. So the general solution is \( u = 2k\pi \), where k is an integer. But the example solution input is \( 3\pi/4, \pi \), which suggests that maybe the original equation was different. But according to the given equation, the solution is \( u = 2k\pi \).

Wait, maybe the user made a typo, and the equation is \( 2\cos^2 u - 7\cos u + 3 = 0 \). Let's solve that. Let \( t = \cos u \), then \( 2t^2 -7t +3 = 0 \). Factoring: (2t -1)(t -3) = 0. So t=1/2 or t=3 (invalid). So \( \cos u = 1/2 \), so \( u = \frac{\pi}{3} + 2k\pi \) or \( u = \frac{5\pi}{3} + 2k\pi \). But since the user's equation is 5, we have to go with that. So the solution is \( u = 2k\pi \), where k is an integer.

But let's confirm with the original equation. For \( u = 2k\pi \), \( \cos u = 1 \), so \( 2(1)^2 -7(1) +5 = 2 -7 +5 = 0 \). Correct. So the solution is \( u = 2k\pi \), where k is an integer.

Answer:

\( u = 2k\pi \) (where \( k \) is an integer)