How do I solve this trigonometric equation?

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Thread starter 2 months ago
#1
3.5 + 7sin( π (t+2)/6)) = 0
I got that π (t+2)/6 should be either -0.524 + n*2π or 3.665 (7π/6) + n*2π but apparently it's all wrong from the very start and it should be 7π/6 + n*2π or 11π/6 + n*2π. Why is -0.524 not an option here?
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2 months ago
#2
(Original post by Nothinghere21)
3.5 + 7sin( π (t+2)/6)) = 0
I got that π (t+2)/6 should be either -0.524 + n*2π or 3.665 (7π/6) + n*2π but apparently it's all wrong from the very start and it should be 7π/6 + n*2π or 11π/6 + n*2π. Why is -0.524 not an option here?
Isnt that just 11pi/6?
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Thread starter 2 months ago
#3
(Original post by mqb2766)
Isnt that just 11pi/6?
Calculator says 11pi/6 is 5.759..., not -0.524...
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2 months ago
#4
(Original post by Nothinghere21)
Calculator says 11pi/6 is 5.759..., not -0.524...
Sure, but its mod 2pi.
Just think of a circle, the angle repeats every 2pi and if you go in the opposite direction ...
Also note -0.524 = -pi/6
Last edited by mqb2766; 2 months ago
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Thread starter 2 months ago
#5
(Original post by mqb2766)
Sure, but its mod 2pi.
Just think of a circle, the angle repeats every 2pi and if you go in the opposite direction ...
Also note -0.524 = -pi/6
But I don't understand why -pi/6 isn't an option. It's literally arcsin(-0.5), so why complicate it by changing it into 11pi/6?
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2 months ago
#6
(Original post by Nothinghere21)
But I don't understand why -pi/6 isn't an option. It's literally arcsin(-0.5), so why complicate it by changing it into 11pi/6?
Generally the principal solutions are in the range [0,2pi] or [-pi,pi]. If it doesn't state in the question, -pi/6 would be acceptable. You'd probably be expected to know arcsin(1/2) = pi/6 and give an exact solution.

11π/6 + n*2π
with n=-1 gives your principal solution.
Last edited by mqb2766; 2 months ago
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