Trying to solve the trig equation <msqrt> 3 + 4 cos 2 </

orlovskihmw 2022-07-09 Answered
Trying to solve the trig equation 3 + 4 cos 2 ( x ) = sin ( x ) 3 + 3 cos ( x )
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Answers (1)

Leslie Rollins
Answered 2022-07-10 Author has 25 answers
My solution goes like this
{ 3 + 4 cos 2 ( x ) = sin 2 ( x ) 3 + 6 3 sin ( x ) cos ( x ) + 9 cos 2 ( x ) sin ( x ) 3 + 3 cos ( x ) 0
3 ( sin 2 ( x ) + cos 2 ( x ) ) + 4 cos 2 ( x ) = sin 2 ( x ) 3 + 6 3 sin ( x ) cos ( x ) + 9 cos 2 ( x )
2 cos 2 ( x ) + sin 2 ( x ) 3 3 sin 2 ( x ) + 6 3 sin ( x ) cos ( x ) = 0
I multiply by 3 and divide by cos 2 ( x ):
8 tan 2 ( x ) 6 3 tan ( x ) 6 = 0
Let t = tan ( x ), then
4 t 2 3 3 t 3 = 0
t = 3 3 ± ( 3 3 ) 2 4 4 ( 3 ) 2 4 = 3 3 ± 5 3 8 = ?
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Uncertain how to solve this trigonometric equation
I am currently attempting to discover how to find the general solutions to
3 tan 2 x = 2 tan x + 3
The given solutions are x = π 3 + π k , 5 π 6 + π k
To solve this equation I removed the square root from 3 tan 2 x leaving me with tan x. Then I subtracted 2 tan x from 3 tan x leaving me with tan x = 3
Should I then not solve tan x = 3 for the generalized solutions?
This would give x = π 3 and x = 2 π 3 which would then be generalized to π 3 + π k and 2 π 3 + π k, which doesn't agree with the given answers.

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