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3sin2x=2cos^2x+1

Guest May 31, 2015

Best Answer 

 #2
avatar+889 
+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

$$\cos2A=2\cos^{2}A-1$$,   from which   $$2\cos^{2}A=\cos2A+1.$$

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

That gets you

$$3\sin2x= \cos2x+2,$$

which is rewritten as

$$3\sin2x-\cos2x = 2.$$

Now consider the trig identity

$$R\sin(2x-\alpha)=R\sin2x\cos\alpha-R\cos2x\sin\alpha.$$

Comparing with the LHS of the equation,

$$R\cos\alpha=3, \text{ and }R\sin\alpha=1,$$

so, squaring and adding, $$R^{2}=3^{2}+1^{2}=10, \text{ so }R=\sqrt{10},$$

and dividing, $$\tan\alpha=1/3.$$

That gets us

$$\sqrt{10}\sin(2x-\alpha)=2,$$

$$\sin(2x-\alpha)=2/\sqrt{10},\quad 2x-\alpha=\sin^{-1}(2/\sqrt{10}),$$

$$2x-\alpha = 39.23,\quad 140.77,\quad 399.23,\quad 500.77, \dots \text{ deg},$$

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

$$x = 28.83, \quad 79.60, \quad 208.83, \quad 259.60, \dots \text{ deg.}$$

Bertie  Jun 1, 2015
 #1
avatar+88871 
+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°

 

 

CPhill  May 31, 2015
 #2
avatar+889 
+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

$$\cos2A=2\cos^{2}A-1$$,   from which   $$2\cos^{2}A=\cos2A+1.$$

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

That gets you

$$3\sin2x= \cos2x+2,$$

which is rewritten as

$$3\sin2x-\cos2x = 2.$$

Now consider the trig identity

$$R\sin(2x-\alpha)=R\sin2x\cos\alpha-R\cos2x\sin\alpha.$$

Comparing with the LHS of the equation,

$$R\cos\alpha=3, \text{ and }R\sin\alpha=1,$$

so, squaring and adding, $$R^{2}=3^{2}+1^{2}=10, \text{ so }R=\sqrt{10},$$

and dividing, $$\tan\alpha=1/3.$$

That gets us

$$\sqrt{10}\sin(2x-\alpha)=2,$$

$$\sin(2x-\alpha)=2/\sqrt{10},\quad 2x-\alpha=\sin^{-1}(2/\sqrt{10}),$$

$$2x-\alpha = 39.23,\quad 140.77,\quad 399.23,\quad 500.77, \dots \text{ deg},$$

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

$$x = 28.83, \quad 79.60, \quad 208.83, \quad 259.60, \dots \text{ deg.}$$

Bertie  Jun 1, 2015
 #3
avatar+88871 
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".......

 

 

CPhill  Jun 1, 2015
 #4
avatar+93311 
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.

Melody  Jun 1, 2015

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