How to find the trigonometric equation for $3 (cos x - 2 {sin}^{2} x) = cos x - 2$ $3(cos x - 2sin^2 x) = cos x - 2$ ?

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Answer 1 · 2018-08-07T02:23:13

$2 cos x - 3 cos 2 x = 1$ $2 cos x - 3 cos 2x = 1$

$3 (cos x - 2 {sin}^{2} x) = cos x - 2$ $3(cos x - 2 sin^2 x) = cos x - 2$

$1 - 2 {sin}^{2} x = cos 2 x$ $1 - 2sin^2 x = cos 2x$

$:. 3 (cos x - cos 2x -1) = cos x - 2$

$3 cos x - 3 cos 2x - 3 = cos x - 2$

$3 cos x - cos x - 3 cos 2x = 3 - 2$

$2 cos x - 3 cos 2x = 1$

Is this your requirement?

Answer 2 · 2018-08-07T05:06:30

$(cos x + 1 )(cos x - 2/3 ) = 0$
$rArr x = { ( 2 k + 1 ) pi}, { 2kpi +- cos^(-1)( 2/3 )}$ ,
$k = 0, +-1, +-2, +-3, ...$ -

Using $sin^2x = 1- cos^2x and c = cos x$ ,

$6c^2 + 2 c - 4 = 2 ( 3 c^2 + c - 2 ) = 0$

#rArr (cos x+1)( cos x - 2/3 ) = 0

How to find the trigonometric equation for 3(cos x - 2sin^2 x) = cos x - 23(cosx−2sin2x)=cosx−23(cos x - 2sin^2 x) = cos x - 2 ?