How do you find all values of x in the interval [0, 2pi] in the equation $6cos^2 x - sin x - 4 = 0$ ?

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Answer 1 · 2016-04-18T17:55:25

$S={2.6180, 0.5236,3.87,5.5535}$

Use the Property $cos^2x+sin^2x=1$

$6(1-sin^2x)-sinx-4=0$

$6-6sin^2x-sinx-4=0$

$6sin^2x+sinx-2=0$

$(2sinx-1)(3sinx+2)=0$

$2sinx-1=0 or 3sinx+2=0$

$2sinx=1 or 3sinx=-2$

$sinx=1/2 or sinx =-2/3$

$x=sin^-1 (1/2) or x=sin^-1 (-2/3)$

$x=pi/6 +2pin, (5pi)/6 +2pin or x=-0.7297...+2pin, -2.4118...+2pin$

$n=0, x=2.6180, 0.5236,-0.7297, -2.4118$

$n=1, x=6.8067, 8.9012, 5.5535, 3.87$

$n=2, x=13.0899, 15.1844, 11.8366, 10.1545->$ all greater than $2pi$ so we stop here

$S={2.6180, 0.5236,3.87,5.5535}$

How do you find all values of x in the interval [0, 2pi] in the equation 6cos^2 x - sin x - 4 = 06cos^2 x - sin x - 4 = 0?