Find the solution (2−(√3)) cos 2x + sin 2x = 1


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Question Number 134705 by bramlexs22 last updated on 06/Mar/21

$Find the solution$ $(2 - \sqrt{3}) \cos 2 x + \sin 2 x = 1$
	Answered by john_santu last updated on 06/Mar/21
	$We can setting → { ((X=cos 2x)),((Y=sin 2x)) :} so the equation becomes { (((2−(√3))X+Y = 1)),((X^2 +Y^( 2) = 1)) :} substituting Y = 1−(2−(√3))X into the second equation gives X^2 +1−2(2−(√3))X+(7−4(√3))X^2 = 1 which has solution X= 0 or X= (1/2) for X = 0 gives Y = 1 and for X=(1/2) gives Y = ((√3)/2) . Now solve the equations { ((cos 2x = 0)),((sin 2x = 1)) :} and { ((cos 2x=(1/2))),((sin 2x=((√3)/2))) :} The solution { ((x=(π/4)+nπ)),((x= (π/6)+nπ)) :} n∈Z$
	$W e c a n s e t t i n g \to {\begin{cases} X = \cos 2 x \\ Y = \sin 2 x \end{cases}$ $s o t h e e q u a t i o n b e c o m e s {\begin{cases} (2 - \sqrt{3}) X + Y = 1 \\ X^{2} + Y^{2} = 1 \end{cases}$ $s u b s t i t u t i n g Y = 1 - (2 - \sqrt{3}) X i n t o$ $t h e s e c o n d e q u a t i o n g i v e s$ $X^{2} + 1 - 2 (2 - \sqrt{3}) X + (7 - 4 \sqrt{3}) X^{2} = 1$ $w h i c h h a s s o l u t i o n X = 0 o r X = \frac{1}{2}$ $f o r X = 0 g i v e s Y = 1 a n d f o r X = \frac{1}{2}$ $g i v e s Y = \frac{\sqrt{3}}{2} . N o w s o l v e t h e e q u a t i o n s$ ${\begin{cases} \cos 2 x = 0 \\ \sin 2 x = 1 \end{cases} a n d {\begin{cases} \cos 2 x = \frac{1}{2} \\ \sin 2 x = \frac{\sqrt{3}}{2} \end{cases}$ $T h e s o l u t i o n {\begin{cases} x = \frac{π}{4} + n π \\ x = \frac{π}{6} + n π \end{cases} n \in Z$
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