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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) + 2sin2(x) =0
Thanks.
Re: Mathematicians, Check In Here! by Martinez39s(m): 4:06am On Feb 14
geraldinai:
Mathematicians in the house, please I need help with this.
Solve for x
3cos(x) + 2sin2(x) =0
Thanks.
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

wink

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Re: Mathematicians, Check In Here! by geraldinai(f): 11:41pm On Feb 14
Martinez39s:
SOLUTION
3Cos x + 2Sin² x = 0
Given that Sin² x = 1 – Cos² x


–2Cos² x + 3Cos x + 2 = 0 is an equal quadrant in form. It is quadratic in Cos x. By substituting u for Cos x, we have


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

wink




Wow! Well explained.
Thanks bro

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