The general solution of equation sinx+sin5x=sin2x+sin4x is?

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The general solution of equation sinx+sin5x=sin2x+sin4x is?

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Answer 1 · 2018-06-22T03:37:34

f(x) = sin x + 5x = sin 2x + sin 4x
Use trig identity:
$sin a + sin b = 2 sin (\frac{a + b}{2}) cos (\frac{a - b}{2}$ $sin a + sin b = 2sin ((a + b)/2)cos ((a - b)/2$
Left side:
sin x + sin 5x = 2sin 3x.cos 2x
Right side:
sin 2x + sin 4x = 2sin 3x.cos x
f(x) = 2sin 3x.cos 2x - 2sin 3x.cos x = 0
(2sin 3x)(cos 2x - cos x) = 0
Either factor should be zero.
a. sin 3x = 0
$x = 2 k π$ $x = 2kpi$ , and $x = π + 2 k π = (2 k + 1) π$ $x = pi + 2kpi = (2k+1)pi$ , and $x = 2 k π$ $x = 2kpi$
b. cos 2x - cos x = 0
Reminder of trig identity:
$cos a - cos b = - 2 sin (\frac{a + b}{2}) . sin (\frac{a - b}{2})$ $cos a - cos b = - 2sin ((a + b)/2).sin ((a - b)/2)$
In this case:
$cos 2 x - cos x = - 2 sin (3 \frac{x}{2}) . sin (\frac{x}{2})$ $cos 2x - cos x = - 2sin (3x/2).sin (x/2)$
a. $sin (3 \frac{x}{2}) = 0$ $sin (3x/2) = 0$ -->
$(3 \frac{x}{2}) = 2 k π$ $(3x/2) = 2kpi$ --> $x = \frac{4 k π}{3}$ $x = (4kpi)/3$
$(3 \frac{x}{2}) = π + 2 k π = (2 k + 1) π$ $(3x/2) = pi + 2kpi = (2k + 1)pi$ --> $x = (2 k + 1) (2 \frac{π}{3})$ $x = (2k + 1)(2pi/3)$
b. $sin (\frac{x}{2}) = 0$ $sin (x/2) = 0$
$\frac{x}{2} = 2 k π$ $x/2 = 2kpi$ --> $x = 4 k π = 2 n π$ $x = 4kpi = 2npi$
$\frac{x}{2} = (2 k + 1) π$ $x/2 = (2k +1)pi$ --> $x = (2 k + 1) 2 π = 2 n π$ $x = (2k +1)2pi = 2npi$

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The general solution of equation sinx+sin5x=sin2x+sin4x is?

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