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$$~~~~~~6 \sin^2 x + 4 \cos^2 x = 5 \tan x \\\\\implies 3 \cdot 2\sin^2 x + 2 \cdot 2 \cos^2 x = 5 \tan x\\ \\\implies 3(1- \cos 2x ) + 2( 1 + \cos 2x) = 5 \tan x\\\\\implies 3 - 3 \cos 2x +2 +2 \cos 2x = 5 \tan x\\\\\implies 5- \cos 2x = 5 \tan x\\ \\\implies 5 - \left(\dfrac{1- \tan^2 x}{ 1+ \tan^2 x} \right)=5 \tan x\\\\\implies \dfrac{5+5 \tan^2x -1 + \tan^2 x}{1+ \tan^2 x} = 5 \tan x\\\\\implies 4 + 6\tan^2 x = 5( \tan x)(1+ \tan^2 x)\\\\\implies 4 +6\tan^2 x = 5 \tan x + 5 \tan^3 x$$

$$\implies 5\tan^3 x -6 \tan^2 x+5 \tan x-4=0\\\\\implies 5 \tan^3 x - 5 \tan^2 x - \tan^2x + \tan x+ 4\tan x -4=0 \\\\\implies 5\tan^2 x(\tan x-1) -\tan x(\tan x-1) +4(\tan x-1)=0\\\\\implies ( \tan x - 1)( 5 \tan^2 x- \tan x +4) =0\\\\\implies \tan x =1\\\\\implies x = n \pi+ \dfrac{\pi}4$$
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$$6 sin² x + 4 cos²x = Stanx$$

$$Correct option is B) Given, 4cos2x+6sin2x=5$$

$$⇒4cos2x+4sin2x+2sin2x=5 ⇒2sin2x=1⇒sin2x=21=(21)2 ⇒x=nπ±4π$$

Solve any question of Trigonometric Functions with:-
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