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Mathematicians, Check In Here! by geraldinai(f): 1:03am On Feb 14 |
Mathematicians in the house, please I need help with this. Solve for x 3cos(x) + 2sin^{2}(x) =0 Thanks. |
Re: Mathematicians, Check In Here! by Martinez39s(m): 4:06am On Feb 14 |
geraldinai:SOLUTION 3Cos x + 2Sin² x = 0 Given that Sin² x = 1 – Cos² x 3Cos x + 2Sin² x = 3Cosx + 2(1 – Cos² x) = –2Cos² x + 3Cos x + 2 –2Cos² x + 3Cos x + 2 = 0 is an equation quadrant in form. It is quadratic in Cos x. By substituting u for Cos x, we have –2u²+ 3u + 2 = 0 The quadratic formula yields two answers: u = –0.5 & u = 2. u = 2 is not in the range of the cosine function so we discard it. The solutions you seek are solutions of the equation Cos x = –0.5 If n is any integer then the solutions are x = 120° + n360° & x = 240° + n360° In radians, x = (2π/3) + 2πn & x = (4π/3) + 2πn 4 Likes |
Re: Mathematicians, Check In Here! by geraldinai(f): 11:41pm On Feb 14 |
Martinez39s: Wow! Well explained. Thanks bro |
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