How do I solve 1-sinx=3cosxfor [0, 2)? Thank you!

Question

$1-sinx=3cosx$ for $[0,2pi)$

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Answer 1 · 2018-02-09T02:03:37

$x=pi/2,-0.9273..+2pi$

$1-sinx=3cosx$

$(1-sinx)^2=(3cosx)^2$

$1-2sinx+sin^2x=9cos^2x$

$1-2sinx+sin^2x=9(1-sin^2x)$

$1-2sinx+sin^2x=9-9sin^2x$

$-8-2sinx+10sin^2x=0$

Let $u=sinx$ :

$10u^2-2u-8=0$

$5u^2-u-4=0$

$(u-1)(5u+4)=0$

$u=1,-4/5$

Plug $sinx$ back in for $u$ :

$sinx=1, -4/5$

$x=pi/2,sin^-1(-4/5)~~-0.9273..$

To get the second number in the wanted range, just add $2pi$ :

$x=pi/2,-0.9273..+2pi$