Solve the following equations: (ix) sin 2x – sin 4x + sin 6x = 0

Home » Solve the following equations: (ix) sin 2x – sin 4x + sin 6x = 0

Solve the following equations: (ix) sin 2x – sin 4x + sin 6x = 0

$\left( \mathbf{ix} \right)~sin\text{ }2x\text{ }\text{ }sin\text{ }4x\text{ }+\text{ }sin\text{ }6x\text{ }=\text{ }0$

Or,

$sin\text{ }2x\text{ }\text{ }sin\text{ }4x\text{ }+\text{ }sin\text{ }6x\text{ }=\text{ }0$

using transformation formula,

$\text{ }sin\text{ }4x\text{ }+\text{ }sin\text{ }6x\text{ }+\text{ }sin\text{ }2x\text{ }=\text{ }0$

By using the formula,

$sin\text{ }A\text{ }+\text{ }sin\text{ }B\text{ }=\text{ }2\text{ }sin\text{ }\left( A+B \right)/2\text{ }cos\text{ }\left( A-B \right)/2$

we get,

$\text{ }sin\text{ }4x\text{ }+\text{ }2\text{ }sin\text{ }\left( 6x+2x \right)/2\text{ }cos\text{ }\left( 6x-2x \right)/2\text{ }=\text{ }0$

$\text{ }sin\text{ }4x\text{ }+\text{ }2\text{ }sin\text{ }4x\text{ }cos\text{ }2x\text{ }=\text{ }0$

Or,

$Sin\text{ }4x\text{ }\left( 2\text{ }cos\text{ }2x\text{ }\text{ }1 \right)\text{ }=\text{ }0$

So,

$Sin\text{ }4x\text{ }=\text{ }0\text{ }or\text{ }2\text{ }cos\text{ }2x\text{ }\text{ }1\text{ }=\text{ }0$

$Sin\text{ }4x\text{ }=\text{ }sin\text{ }0\text{ }or\text{ }cos\text{ }2x\text{ }=\text{ }{\scriptscriptstyle 1\!/\!{ }_2}$

Or,

$Sin\text{ }4x\text{ }=\text{ }sin\text{ }0\text{ }or\text{ }cos\text{ }2x\text{ }=\text{ }\pi /3$

$4x\text{ }=\text{ }n\pi \text{ }or\text{ }2x\text{ }=\text{ }2m\pi \text{ }\pm \text{ }\pi /3$

Or,

$x\text{ }=\text{ }n\pi /4\text{ }or\text{ }x\text{ }=\text{ }m\pi \text{ }\pm \text{ }\pi /6$

∴ the general solution is

$x\text{ }=\text{ }n\pi /4\text{ }or\text{ }m\pi \text{ }\pm \text{ }\pi /6,$

where m, n ϵ Z.

i

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