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If
5sin 2x +4 = 10 sin ²x, show
that tanx = 2 or tan x =-1/3

Sagot :

[tex]~~~~~5\sin 2x +4 =10 \sin^2 x\\\\\implies 5\cdot 2\sin x \cos x+4 =10 \sin^2 x\\\\\implies 10 \sin x \cos x +4 = 10\sin^2 x\\\\\implies 10\tan x + 4\sec^2 x = 10\tan^2 x~~~~~;[\text{Divide both sides by}~ \cos^2 x]\\\\\implies 10\tan^2 x -10 \tan x -4(1+ \tan^2 x)=0\\\\\implies 10\tan^2 x -10 \tan x -4 -4\tan^2 x =0\\\\\implies 6 \tan^2 x -10 \tan x -4 =0\\\\\implies 3 \tan^2 x -5 \tan x -2 =0~~~~~;[\text{Divide both sides by 2}]\\\\\implies 3\tan^2 x - 6\tan x +\tan x -2 =0\\\\[/tex]

[tex]\implies 3 \tan x( \tan x -2) + (\tan x -2) = 0\\\\\implies (\tan x -2) ( 3 \tan x +1) = 0\\\\\implies \tan x = 2 ~~ \text{or}~~ \tan x = - \dfrac 13\\\\\text{Showed.}[/tex]