How do you find a numerical value of one trigonometric function of x given $cos^2x+2sinx-2=0$ ?

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Answer 1 · 2016-11-14T19:36:28

$:.x=pi/2 + 2pin$

$cos^2x+2sinx-2=0$

$(1-sin^2x)+2sinx-2=0$

$-sin^2x+2sinx-1=0$

$sin^2x-2sinx+1=0$

Now factor.

$(sinx-1)(sinx-1)=0$

$(sinx-1)^2=0$

$sinx-1=0$

$sinx=1$

$x=sin^-1 1$

$:.x=pi/2 + 2pin$

How do you find a numerical value of one trigonometric function of x given cos^2x+2sinx-2=0cos^2x+2sinx-2=0?