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Solve the equation (cosx+cos3x+cos5x)/(sinx+sin3x+sin5x)=square root of 3.

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sixpenpencil | Student, Grade 10 | eNoter

Posted December 23, 2010 at 1:30 AM via web

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Solve the equation (cosx+cos3x+cos5x)/(sinx+sin3x+sin5x)=square root of 3.

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justaguide | College Teacher | (Level 2) Distinguished Educator

Posted December 23, 2010 at 1:56 AM (Answer #1)

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We have to solve the equation (cos x + cos 3x + cos 5x) / (sin x + sin 3x + sin 5x) = sqrt 3.

Now we know that cos x + cos y= 2cos[x+y)/2]*cos [(x-y)/2]

=>cos x + cos (5x)= 2 cos[(x+5x)/2]*cos[(x-5x)/2]

=> cos x + cos (5x)= 2cos(3x)cos(2x)

=> (cos x + cos 3x + cos 5x) = 2cos(3x)cos(2x) + cos 3x

=> cos 3x( 2cos 2x +1)

sin A + sin B = 2 sin [ (A + B) / 2 ] cos [ (A - B) / 2 ]

=> sin x + sin 5x = 2 sin 3x * cos 2x

sin x + sin3x + sin5x = 2 sin 3x * cos 2x + sin 3x

= sin 3x( 2cos 2x + 1)

cos 3x( 2cos 2x +1)/ sin 3x( 2cos 2x + 1)

= 1/ tan 3x = sqrt 3

tan 3x = 1/ sqrt 3

3x = arc tan (1/ sqrt 3)

=> 3x = 30 degrees

=> x=  10 degrees.

Therefore x = 10 degrees.

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neela | High School Teacher | Valedictorian

Posted December 23, 2010 at 3:01 AM (Answer #3)

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cosx+cos5x = 2cos((x+5x)/2)*cos((x-5x)/2) = 2cos3xcos2x, as cos(-A) cosA.

Therefore cox+cos3x+cos5x = 2cos3x+cos2x+cos3x = cos3x{2cos2x + 1)...(1)

Similarly sinx +sin5x =  2sin(x+5x)/2 *cos(x-5x)/2 = 2sin3x cos2x.

Therefore sinx+sin3x+sin5x = 2sin33x*cos2x+sin3x= sin3x(2cos2x + 1)....(2)

Therefore (cosx+cos3x+cos5x)/(sinx+sin3x+sin5x) = cos3x{2cos2x + 1)/sin3x(2cos2x + 1) = cos3x/sin3x = sqrt3.

=> cot3x = sqrt3.

 3x = pi/3 , or 3x =  5pi/3 ,

Or x= pi/9. Or x = 5pi/9.

 

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