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General Solution of Trigonometric Equation

Prove that: ...

Question

Prove that: $\frac{cos 5 x + cos 4 x}{2 cos 3 x - 1} = cos x + cos 2 x$ .

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Solution

LHS $= - \frac{cos 5 x + cos 4 x}{1 - 2 cos 3 x}$
$c o s C + c o s D = 2 cos (\frac{C + D}{2}) cos (\frac{C - D}{2})$
$= - ⎡ ⎢ ⎢ ⎢ ⎢ ⎣ \frac{2 cos (\frac{9 x}{2}) cos (\frac{x}{2})}{1 - 2 cos 3 x} ⎤ ⎥ ⎥ ⎥ ⎥ ⎦$
Multiply and divide by $cos (\frac{3 x}{2})$
$= - ⎡ ⎢ ⎢ ⎢ ⎢ ⎣ \frac{2 cos (\frac{9 x}{2}) cos (\frac{x}{2}) cos (\frac{3 x}{2})}{cos (\frac{3 x}{2}) - 2 cos 3 x cos \frac{3 x}{2}} ⎤ ⎥ ⎥ ⎥ ⎥ ⎦$
$= - ⎡ ⎢ ⎢ ⎢ ⎣ \frac{2 cos \frac{9 x}{2} \times cos \frac{x}{2} \times cos \frac{3 x}{2}}{cos \frac{3 x}{2} - cos \frac{9 x}{2} - cos \frac{3 x}{2}} ⎤ ⎥ ⎥ ⎥ ⎦$
$∵ 2 c o s A c o s B = cos (A + B) + cos (A - B)$
$= - ⎡ ⎢ ⎢ ⎢ ⎣ \frac{2 cos \frac{9 x}{2} cos \frac{x}{2} cos \frac{3 x}{2}}{- cos \frac{9 x}{2}} ⎤ ⎥ ⎥ ⎥ ⎦ = 2 cos \frac{x}{2} cos \frac{3 x}{2}$
$= cos (2 x) + c o s x$ . RHS

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General Solution of Trigonometric Equation

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