Question

Does 2sinxtanx = 3 ===
2-2cos^2x over 2cosx=3?

Answer 1

\[{2-2\cos^2x \over 2cosx}=3\]

Answer 2

You want to find the solution? or just proving?

Answer 3

prove and find the solution. But is 2sinxtanx equivalent to what I wrote?

Answer 4

the answer to your question is if they're equivalent is no. cross out the leading 2 and it will be.

Answer 5

but that's not the route i'd take...

Answer 6

all yours sam..

Answer 7

\[\frac{2(1-\cos^{2}x)}{2cosx}=3\]
\[\frac{(1-\cos^{2}x)}{cosx}=3\]

\[1-\cos^{2}x=3cosx\]
\[-\cos^{2}x-3cosx+1=0\]

Answer 8

No whole number solutions

Answer 9

Hmm Ok... But there should be... I think

Answer 10

Solution for that is
\[x=\sin ^{-1}\left(\sqrt{\frac{3}{2} \left(\sqrt{13}-3\right)}\right)\approx 1.26319\]
and
\[x=-\sin ^{-1}\left(\sqrt{\frac{3}{2} \left(\sqrt{13}-3\right)}\right)\approx -1.26319\]

Answer 11

Won't it be

2sinxtanx = 3
=> 2(sin^2x)/cosx = 3
=> 2(1 - cos^2x)/cosx = 3
=> 2 - 2cos^2x = 3cosx
=> 2cos^2x + 3cosx - 2 = 0
=> (2cosx - 1)(cosx + 2) = 0

cosx = 1/2, cosx = -2 => Impossible

=> x = Pi/3, 2Pi - Pi/3 = 5Pi/3

Answer 12

The "2" canceled from start ,
the furthest you can get is
sin(x) tan(x)=3

Answer 13

2sinxtanx is not the same as 2-2cos^x over 2cosx

Answer 14

Ok. Is it the same as 2-2cos^2x over cosx?

Answer 15

yes

Answer 16

Thank you :)

Answer 17

You're welcome.. I didn't help anyway...

Answer 18

@c How do you get from the ranges 0<x<360 with the answer
cosx=1/2?

The first is 60, and then what?

Answer 19

@Callisto

Answer 20

Very ugly drawing sorry.
But from the graph, we can know there are 2 solutions for cosx =0.5
as you know the first one is 60
the second solution is in quadrant IV , which gives positive value for cosx (i meant it is like the quadrant I)
and from cos (360 -x) = cos x, you can get the 2nd angle by 360 -x =60
Hmm.. is that your question?

Answer 21

Yes, thank you :)

Answer 22

you're welcome!