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How do you solve $4 cos x \cdot cos 2 x \cdot cos 4 x = sin 8 x$ $4\cos x \cdot \cos 2x \cdot \cos 4x = \sin 8x$ ?

1 Answer

Apr 5, 2018

$x = 30$ $x=30$

Explanation:

$4 cos (x) \cdot cos (2 x) \cdot cos (4 x) = sin (8 x)$ $4cos(x)*cos(2x)*cos(4x)=sin(8x)$

$sin 2 (A) = 2 sin (A) cos (A)$ $sin2(A)=2sin(A) cos(A)$ (Identity)

$sin (8 x) = sin (2 \times 4 x)$ $sin(8x)=sin(2xx4x)$ $(4 x = A)$ $(4x=A)$

$sin (2 \times 4 x) = 2 cos (4 x) sin (4 x)$ $sin(2xx4x)=2cos(4x)sin(4x)$

$4 cos (x) \cdot cos (2 x) \cdot cos (4 x) = 2 cos (4 x) \cdot sin (4 x)$ $4cos(x)*cos(2x)*cos(4x)=2cos(4x)*sin(4x)$

Divide by $cos (4 x)$ $cos(4x)$ both sides

$4 cos (x) \cdot cos (2 x) = 2 sin (4 x)$ $4cos(x)*cos(2x)=2sin(4x)$

Again

$sin (4 x) = sin (2 \times 2 x)$ $sin(4x)=sin(2xx2x)$ $(2 x = A)$ $(2x=A)$

$sin (2 \times 2 x) = 2 cos (2 x) sin (2 x)$ $sin(2xx2x)=2cos(2x)sin(2x)$

$4 cos (x) \cdot cos (2 x) = 4 cos (2 x) \cdot sin (2 x)$ $4cos(x)*cos(2x)=4cos(2x)*sin(2x)$

Divide by $cos (2 x)$ $cos(2x)$ both sides

$4 cos (x) = 4 sin (2 x)$ $4cos(x)=4sin(2x)$ and divide by 4

$cos (x) = sin (2 x)$ $cos(x)=sin(2x)$

$sin (2 x) = sin (2 \times x)$ $sin(2x)=sin(2xxx)$ $(x = A)$ $(x=A)$

$sin (2 \times x) = 2 cos (x) \cdot sin (x)$ $sin(2xxx)=2cos(x)*sin(x)$

$cos (x) = 2 cos (x) \cdot sin (x)$ $cos(x)=2cos(x)*sin(x)$

So, $sin (x) = \frac{1}{2}$ $sin(x)=1/2$

$x = 30$ $x=30$

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