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3sin2x=2cos^2x+1

 May 31, 2015

Best Answer 

 #2
avatar+893 
+13

The graphical method is fine if you have access to a suitable plotter, if you haven't, having only a pocket calculator, your stuck.

Here's the mathematicians method of solution.

Begin with a substitution using the trig identity

cos2A=2cos2A1,   from which   2cos2A=cos2A+1.

(in order to get the angles on both sides the same).

That gets you

3sin2x=cos2x+2,

which is rewritten as

3sin2xcos2x=2.

Now consider the trig identity

Rsin(2xα)=Rsin2xcosαRcos2xsinα.

Comparing with the LHS of the equation,

Rcosα=3, and Rsinα=1,

so, squaring and adding, R2=32+12=10, so R=10,

and dividing, tanα=1/3.

That gets us

10sin(2xα)=2,

sin(2xα)=2/10,2xα=sin1(2/10),

2xα=39.23,140.77,399.23,500.77, deg,

and with alpha = 18.43 deg, that leads to (0 - 360 deg, 2dp),

x=28.83,79.60,208.83,259.60, deg.

 Jun 1, 2015
 #1
avatar+130458 
+10

3sin2x=2cos^2x+1   

 

Here's a graphical solution.........https://www.desmos.com/calculator/ynevyjlzji

 

There are four solutions on [0, 360] degrees......these occur at about 28.8°, 79.6°, 208.8° and 259.6°

 

 

 May 31, 2015
 #2
avatar+893 
+13
Best Answer

The graphical method is fine if you have access to a suitable plotter, if you haven't, having only a pocket calculator, your stuck.

Here's the mathematicians method of solution.

Begin with a substitution using the trig identity

cos2A=2cos2A1,   from which   2cos2A=cos2A+1.

(in order to get the angles on both sides the same).

That gets you

3sin2x=cos2x+2,

which is rewritten as

3sin2xcos2x=2.

Now consider the trig identity

Rsin(2xα)=Rsin2xcosαRcos2xsinα.

Comparing with the LHS of the equation,

Rcosα=3, and Rsinα=1,

so, squaring and adding, R2=32+12=10, so R=10,

and dividing, tanα=1/3.

That gets us

10sin(2xα)=2,

sin(2xα)=2/10,2xα=sin1(2/10),

2xα=39.23,140.77,399.23,500.77, deg,

and with alpha = 18.43 deg, that leads to (0 - 360 deg, 2dp),

x=28.83,79.60,208.83,259.60, deg.

Bertie Jun 1, 2015
 #3
avatar+130458 
0

Thanks, Bertie.....I've seen you use this method before, but I didn't remember it.......I'll have to keep this one in "reserve".......

 

 

 Jun 1, 2015
 #4
avatar+118696 
0

Thanks Bertie, it is always good to see you on the forum :)

 

I would have done it Bertie's way but I always like to see your graphical solutions too Chris.

Sometimes I forget that problems can be tackled graphically too.

 Jun 1, 2015

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