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}#