((JS)/(__00_00__00)) solve the equation 4sin 3x + (1/3)cos 3x = 3


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Question Number 109401 by john santu last updated on 23/Aug/20

$\frac{J S}{__00_00__00}$ $s o l v e t h e e q u a t i o n 4 \sin 3 x + \frac{1}{3} \cos 3 x = 3$
	Answered by mr W last updated on 23/Aug/20

	$12 \sin 3 x + \cos 3 x = 9$ $\frac{12}{\sqrt{12^{2} + 1^{2}}} \sin 3 x + \frac{1}{\sqrt{12^{2} + 1^{2}}} \cos 3 x = \frac{9}{\sqrt{12^{2} + 1^{2}}}$ $\sin α \sin 3 x + \cos α \cos 3 x = \frac{9}{\sqrt{145}}$ $\cos (3 x - α) = \frac{9}{\sqrt{145}}$ $3 x - α = 2 k π \pm \cos^{- 1} \frac{9}{\sqrt{145}}$ $\Rightarrow x = \frac{1}{3} (2 k π + α \pm \cos^{- 1} \frac{9}{\sqrt{145}})$ $\Rightarrow x = \frac{1}{3} (2 k π + \cos^{- 1} \frac{1}{\sqrt{145}} \pm \cos^{- 1} \frac{9}{\sqrt{145}})$
	Answered by ajfour last updated on 24/Aug/20

	$\sin (3 x + ϕ) = \frac{9}{\sqrt{145}}$ $x = \frac{1}{3} (\tan^{- 1} \frac{1}{12} + \tan^{- 1} \frac{9}{8})$ $\tan 3 x = \frac{\frac{1}{12} + \frac{9}{8}}{1 - \frac{9}{12 \times 8}} = \frac{116}{87}$ $x = \frac{n π}{3} + \frac{1}{3} \tan^{- 1} (\frac{116}{87}) .$
	Commented by mr W last updated on 23/Aug/20

	$n i c e s o l u t i o n!$
	Answered by 1549442205PVT last updated on 23/Aug/20
	$Put tan((3x)/2)=y ⇒sin3x=((2y)/(1+y^2 )),cos3x=((1−y^2 )/(1+y^2 )) we get ((8y)/(1+y^2 ))+((1−y^2 )/(3(1+y^2 )))=3 ⇔24y+1−y^2 =9(1+y^2 ) ⇔10y^2 −24y+8=0⇔5y^2 −12y+4=0 Δ′=36−20=16⇒y=((6±4)/5)∈{2,(2/5)} i)For y=2⇔tan((3x)/2)=2⇔((3x)/2)=tan^(−1) 2+kπ ⇔x=(2/3)tan^(−1) 2+((2kπ)/3) ii)For y=(2/5)⇔tan((3x)/2)=(2/5)⇔((3x)/2)=tan^(−1) ((2/5) )+kπ ⇔x=(2/3)tan^(−1) ((2/5))+((2kπ)/3) Thus,the roots of given equation are x∈{(2/3)tan^(−1) 2+((2k𝛑)/3);(2/3)tan^(−1) ((2/5))+((2k𝛑)/3)}$
	$Put \tan \frac{3 x}{2} = y \Rightarrow \sin 3 x = \frac{2 y}{1 + y^{2}}, \cos 3 x = \frac{1 - y^{2}}{1 + y^{2}}$ $we get \frac{8 y}{1 + y^{2}} + \frac{1 - y^{2}}{3 (1 + y^{2})} = 3$ $\Leftrightarrow 24 y + 1 - y^{2} = 9 (1 + y^{2})$ $\Leftrightarrow {10 y}^{2} - 24 y + 8 = 0 \Leftrightarrow {5 y}^{2} - 12 y + 4 = 0$ $Δ^{'} = 36 - 20 = 16 \Rightarrow y = \frac{6 \pm 4}{5} \in {2, \frac{2}{5}}$ $i) For y = 2 \Leftrightarrow \tan \frac{3 x}{2} = 2 \Leftrightarrow \frac{3 x}{2} = \tan^{- 1} 2 + k π$ $\Leftrightarrow x = \frac{2}{3} \tan^{- 1} 2 + \frac{2 k π}{3}$ $ii) For y = \frac{2}{5} \Leftrightarrow \tan \frac{3 x}{2} = \frac{2}{5} \Leftrightarrow \frac{3 x}{2} = \tan^{- 1} (\frac{2}{5}) + k π$ $\Leftrightarrow x = \frac{2}{3} \tan^{- 1} (\frac{2}{5}) + \frac{2 k π}{3}$ $Thus, the roots of given equation are$ $x \in {\frac{2}{3} \tan^{- 1} 2 + \frac{2 k π}{3}; \frac{2}{3} \tan^{- 1} (\frac{2}{5}) + \frac{2 k π}{3}}$
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