find the general solution for 2sin 3x = sin 2x


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Question Number 79969 by Rio Michael last updated on 29/Jan/20

$f i n d t h e g e n e r a l s o l u t i o n f o r$ $2 \sin 3 x = \sin 2 x$
	Answered by behi83417@gmail.com last updated on 29/Jan/20
	$2(3sinx−4sin^3 x)=2sinxcosx ⇒sinx(3−4sin^2 x−cosx)=0 ⇒ { ((sinx=0⇒x=kπ ,k∈Z)),((3−4(1−cos^2 x)−cosx=0)) :} 4cos^2 x−cosx−1=0 ⇒cosx=((1±(√(1+16)))/8)=((1±(√(17)))/8) ⇒ { ((cosx=((1+(√(17)))/8)⇒x=2mπ±cos^(−1) ((1+(√(17)))/8)[m∈Z])),((cosx=((1−(√(17)))/8)⇒x=2nπ±cos^(−1) ((1−(√(17)))/8)[n∈Z])) :}$
	$2 (3 sinx - {4 \sin}^{3} x) = 2 sinxcosx$ $\Rightarrow sinx (3 - {4 \sin}^{2} x - cosx) = 0$ $\Rightarrow {\begin{cases} sinx = 0 \Rightarrow x = k π, k \in Z \\ 3 - 4 (1 - \cos^{2} x) - cosx = 0 \end{cases}$ ${4 \cos}^{2} x - cosx - 1 = 0$ $\Rightarrow cosx = \frac{1 \pm \sqrt{1 + 16}}{8} = \frac{1 \pm \sqrt{17}}{8}$ $\Rightarrow {\begin{cases} cosx = \frac{1 + \sqrt{17}}{8} \Rightarrow x = 2 m π \pm \cos^{- 1} \frac{1 + \sqrt{17}}{8} [m \in Z] \\ cosx = \frac{1 - \sqrt{17}}{8} \Rightarrow x = 2 n π \pm \cos^{- 1} \frac{1 - \sqrt{17}}{8} [n \in Z] \end{cases}$
	Commented by Rio Michael last updated on 30/Jan/20

	$t h a n k s$
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