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How would you solve this trig equation?

 2sin2x+sinx=0 

Where 0<x<360

 Nov 24, 2015

Best Answer 

 #2
avatar+128578 
+15

2sin2x + sinx   =  0         factor out sin x

 

sinx [ 2sinx + 1]  =  0     set both factors  =  0

 

sin x  =  0       and this happens at 180 °  on   0 > x > 360

 

And

 

2sin x  +  1  = 0        subtract 1 from both sides and divide both sides by 2

 

sinx = -1/2      and this happens at 210° and at 330°  on the interval of interest

 

 

 

cool cool cool

 Nov 25, 2015
 #1
avatar+118608 
+10

Hi Steven434, welcome to the forum :)

 

This is NOT a calculus question - it is a trig question ://

 

How would you solve this trig equation?

 2sin2x+sinx=0 

Where 0<x<360

 

\( 2sin2x+sinx=0 \\ 2(sin^2x-cos^2x)+sinx=0\\ 2(sin^2x-[1-sin^2x])+sinx=0\\ 2(sin^2x-1+sin^2x)+sinx=0\\ 2(2sin^2x-1)+sinx=0\\ 4sin^2x-2+sinx=0\\ 4sin^2x+sinx-2=0\\ let\;y=sinx\\ 4y^2+y-2=0 \)

 

4y^2+y-2=0 = {    y=-(((sqrt(33)+1)/8)),       y=((sqrt(33)-1)/8)}

-(((sqrt(33)+1)/8)) = -0.8430703308172536

((sqrt(33)-1)/8) = 0.5930703308172536

 

so

sinx = - 0.8430703308172536    (this will be in 3rd and 4th quadrants)     

              OR         sinx = 0.5930703308172536             (this will be in 1st and 2th quadrants) 

 

 

sinx = 0.8430703308172536                 OR             sinx = 0.5930703308172536

x=asin(0.8430703308172536)

asin(0.8430703308172536) = 57.46577 degrees

asin(-0.8430703308172536) = 180+57.46577     or      360-57.46577       degrees

asin(-0.8430703308172536) = 237.46577     or      302.53423       degrees

asin(0.5930703308172536) = 36.375192227   or    180-36.375192227

asin(0.5930703308172536) = 36.375192227   or    143.6248077  degrees

 

So to 2 decimal places the answers are 

36.38   and    143.62  and     237.47    and      302.53      degrees

 Nov 24, 2015
 #2
avatar+128578 
+15
Best Answer

2sin2x + sinx   =  0         factor out sin x

 

sinx [ 2sinx + 1]  =  0     set both factors  =  0

 

sin x  =  0       and this happens at 180 °  on   0 > x > 360

 

And

 

2sin x  +  1  = 0        subtract 1 from both sides and divide both sides by 2

 

sinx = -1/2      and this happens at 210° and at 330°  on the interval of interest

 

 

 

cool cool cool

CPhill Nov 25, 2015
 #3
avatar+118608 
0

Thanks Chris :)

 

sin2x+sinx=0 

 

that is interesing

When i copied and pasted the question into my post the position of the 2 changed !

I didn't know that happened.  !!

How annoying!

 Nov 25, 2015
 #4
avatar+128578 
+5

Yep....I noticed that the "copy and paste" function can't distinguish between exponents and non-exponential numbers....it treats all of them as non-exponential  [ unless, of course, you use the " ^ "  symbol ]

 

 

 

 

 

cool cool cool

 Nov 25, 2015
 #5
avatar+118608 
0

On the front (question) page of the forum it is 2x as well!  

 

I mean where all the questions are.  :/

 Nov 25, 2015
 #6
avatar+128578 
+5

If this is

 

2sin2x + sinx = 0

 

2*2sinxcosx + sinx  = 0     

 

4sinxconx + sinx  = 0    factor out sinx

 

sinx ( 4cosx + 1)  = 0

 

Then either

 

sin x = 0    and this happpens at 180°

 

Or

 

4cosx +  1 = 0

 

cosx  = -1/4      and this happens at about 104.48°   and at  about 252.52 °

 

Here's the graph.......https://www.desmos.com/calculator/xr4vh9ar1d

 

 

cool cool cool

 Nov 25, 2015

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