Write $sin x + sin 3 x$ $sin x +sin 3x$ as a product. Hence,find all angles $x$ $x$ in the interval $[0, 2 π]$ $[0,2pi]$ that satisfy $sin x + sin 3 x - cos x = 0$ $sin x +sin 3x - cos x =0$ ?

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Answer 1 · 2015-09-01T14:27:54

$x \in {\frac{π}{12}, \frac{5 π}{12}, \frac{π}{2}, \frac{3 π}{2}}$ $x in{pi/12," "(5pi)/12," "pi/2," "(3pi)/2}$

First you should be aware that:
$sin A + sin B \equiv 2 sin (\frac{A + B}{2}) cos (\frac{A - B}{2})$ $sinA+sinB-=2sin((A+B)/2)cos((A-B)/2)$

So using the above result,
$sin x + sin 3 x = 2 sin (\frac{x + 3 x}{2}) cos (\frac{x - 3 x}{2})$ $sinx+sin3x=2sin((x+3x)/2)cos((x-3x)/2)$

$= 2 sin (2 x) cos (- x) = 2 sin (2 x) cos (x)$ $=2sin(2x)cos(-x)=color(blue)(2sin(2x)cos(x))$

Hence,

$sin x + sin 3 x - cos x = 0$ $sinx+sin3x-cosx=0$

Becomes,

$2 sin (2 x) cos (x) - cos (x) = 0$ $2sin(2x)cos(x)-cos(x)=0$

Factor out $cos (x)$ $cos(x)$

$\Rightarrow cos (x) [2 sin (2 x) - 1] = 0$ $=>cos(x)[2sin(2x)-1]=0$

Case 1 :
$cos (x) = 0$ $cos(x)=0$

Thus , $x = \pm \frac{π}{2} + 2 n π$ $x=+-pi/2+2npi" "$ , $ninZZ$

Now we substitute all the values of $n$ that will give us answers in in the interval $[0,2pi]$

$n=0" " =>x=color(orange)(pi/2)$
$n=1" " =>x=-pi/2+2pi=color(orange)((3pi)/2)$

Case 2:
$2sin(2x)-1=0$

$=>sin(2x)=1/2$

$=>2x=pi/6(-1)^n+npi" "$ , $ninZZ$
$=>x=pi/12(-1)^n+npi/2$

Again, now we substitute all the values of $n$ that give us the range we want,

$n=0" " =>x=color(orange)(pi/12)$

$n=1" " =>x=-pi/12+pi/2=color(orange)((5pi)/12)$

Write sin x +sin 3xsinx+sin3xsin x +sin 3x as a product. Hence,find all angles xxx in the interval[0,2pi][0,2π][0,2pi] that satisfy sin x +sin 3x - cos x =0sinx+sin3x−cosx=0sin x +sin 3x - cos x =0?