How do you solve for x if $cos (6x-20) = sin (2x - 10)$ ?

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Answer 1 · 2018-07-24T02:30:19

$x=15$

$sin x =cos (90-x)$
$cosx = sin (90-x)$

$cos (6x-20)=sin(90-(6x-20))=sin(90-6x+20)=sin(110-6x)$

$sin(110-6x)=sin(2x-10)$

$110-6x=2x-10$

$120=8x$

$x=15$

Sub $x=15$ into the equation
LHS:
$cos(6x-20)$
$=cos(6times15-20)$
$=cos(90-20)$
$=cos70$

RHS:
$sin(2x-10)$
$=sin(2times15-10)$
$=sin(30-10)$
$=sin20$
$=cos(90-20)$
$=cos70$
$=LHS$

Therefore, $x=15$ is correct

How do you solve for x if cos (6x-20) = sin (2x - 10)cos (6x-20) = sin (2x - 10)?