Question

If α is the value of $xϵ[0,\pi ]$ satisfying 3 cos x + 3 sin x + sin 3x - cos 3x = 0, then find the value of $\frac{4\alpha }{\pi }$ ?

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Answer 1

We need to simplify the given expression to find the solutions. We observe that the given expression has

sin x + cos x. So if we can express sin 3x — cos 3x in terms of sin x+ cos x and sin x cos x, we can solve it

using the substitution sin x + cos x= t
Consider sin 3x —cos 3x.

$sin3x-cos3x=3sinx-4si{n}^{3}x-4co{s}^{3}x+3cosx$
$=3(sinx+cosx)-4(si{n}^{3}x+co{s}^{3}x)$

We have,$(sinx+cosx{)}^3=si{n}^{3}x+co{s}^{3}x+3sinxcosx(cosx+sinx)$
$\Rightarrow$ $si{n}^{3}x+co{s}^{3}x=(sinx+cosx{)}^3—3sinxcosx(sinx+cosx)$

$\Rightarrow$ we got all the terms in the given expression in terms of (sin x + cos x) and sin x cos x. so we will go for

the substitution sin x+ cos x = t

$\Rightarrow 3cosx+3sinx+sin3x-cos3x=3t+3t-4[t^3-3\frac{t^2-1}{2}t][\because sinxcosx=\frac{t^2-1}{2}]$
$=6t+2t^3-6t$
$=2t^3$
$\Rightarrow 2t^3=0$

$\Rightarrow t=0$

$\Rightarrow sinx+cosx=0$

$\Rightarrow tanx=-1=tan\left(\frac{3\pi }{4}\right)$
=> answer = 3