# trig question help

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#1

is it sinx(7/cosx - 8) = 0
so sin(4a +199) = 0 and
cos(4a+199) = 7/8
the answer is 33 and solving the 2 dont give 33?
Last edited by r.maliak; 3 weeks ago
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3 weeks ago
#2
(Original post by r.maliak)

is it sinx(7/cosx - 8) = 0
so sin(4a +199) = 0 and
cos(4a+199) = 7/8
the answer is 33 and solving the 2 dont give 33?
How did you solve the asin and acos parts?
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#3
(Original post by mqb2766)
How did you solve the asin and acos parts?
sin(4a+199) = 0
sin-1(0) = 0
4a+199 = 0
a = -199/3
0
3 weeks ago
#4
(Original post by r.maliak)
sin(4a+199) = 0
sin-1(0) = 0
4a+199 = 0
a = -199/3
There are multiple solutions for
sin(*) = 0
The +199 is a hint that the obvious one, may not be the correct one.

Similarly for cos(*) solutions
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