Question

cos3x cos cube x + sin 3x sin cube x = 0

Answer 1

Tan3 x.tan^3 x = -1

Friend now which value u want achive

Answer 2

Given:

$\mathbf{\cos 3X\times \cos^{3}X +\sin 3X\times \sin^{3}X =0 }$      ...a)

To Find:

Find general solution of given equation?

Solution:

We know the formula some identity equation and trigonometric formula:

(a-b)³ =a³ -b³ -3×a²×b +3×a×b²     ...1)

$\mathbf{\tan 3X=\dfrac{3\tan X-\tan ^{3}X}{1-3\tan ^{2}X} }$           ...2)

Divide both by $\mathbf{\cos 3X\times \cos^{3}X }$ in above  equation a):

$\mathbf{\dfrac{\cos 3X\times \cos^{3}X}{\cos 3X\times \cos^{3}X} +\dfrac{\sin 3X\times \sin^{3}X}{\cos 3X\times \cos^{3}X} =\dfrac{0}{\cos 3X\times \cos^{3}X} }$

$\mathbf{1 +\tan 3X\times \tan^{3}X =0 }$               ...3)

From equation 2) and equation 3) :

$\mathbf{1+\left ( \dfrac{3\tan X-\tan ^{3}X}{1-3\tan ^{2}X} \right )\times \tan ^{3}X =0 }$

$\mathbf{1+\dfrac{3\tan^{4} X-\tan ^{6}X}{1-3\tan ^{2}X} =0 }$

$\mathbf{\dfrac{1-3\tan ^{2}X+3\tan^{4} X-\tan ^{6}X}{1-3\tan ^{2}X} =0 }$

Above equation can also be written as:

$\mathbf{\dfrac{\tan ^{6}X-1-3\tan^{4} X+3\tan ^{2}X}{-1+3\tan ^{2}X} =0 }$   ...4)

From equation 1) and equation 4), numerator of equation 4) can be written as:

$\mathbf{\dfrac{\left ( \tan ^{2}X -1 \right )^{3}}{-1+3\tan ^{2}X} =0 }$                        ...5)

Above equation only is satisfied when numerator of equation 5) is zero:

$\mathbf{( \tan ^{2}X -1 )^{3}=0 }$

$\mathbf{\tan ^{2}X =1 }$

tan X = ± 1                                    ...6)

If tan X = +1

  $\mathbf{\tan X= \tan \dfrac{\pi }{4} }$

Its general solution is :

  $\mathbf{X = n\pi +\dfrac{\pi }{4}}$                            ...7)

If tan X = -1

  $\mathbf{\tan X= \tan \dfrac{3\pi }{4} }$

Its general solution is :

   $\mathbf{X = n\pi +\dfrac{3\pi }{4}}$                           ...8)

Equation 7) and equation 8) are general solution of given expression$\mathbf{\cos 3X\times \cos^{3}X +\sin 3X\times \sin^{3}X =0 }$.