How do you solve # 2sin^2x - cosx = 1# over the interval 0 to 2pi?

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Answer 1 · 2016-02-14T03:55:27

#x=60^@, 300^@, 180^@" "# or
#x=pi/3, (5pi)/3, pi#

Start from the given #2 sin^2 x - cos x = 1#

note that, #sin^2 x+cos^2 x=1# and #sin^2 x=1-cos^2 x#

and #2 sin^2 x - cos x = 1#

is also #2*(1-cos^2 x)-cos x=1#

and

#2-2cos^2 x-cos x=1#

#2 cos^2 x + cos x-2+1=0#

#2 cos^2 x + cos x-1=0#

use #color(red)("Quadratic Equation")#

#cos x=(-b+-sqrt(b^2-4ac))/(2a)#

#2 cos^2 x + cos x-1=0#

from #2 cos^2 x + 1*cos x-1=0#

Let #a=2#, #b=1#, #c=-1#

#cos x=(-b+-sqrt(b^2-4ac))/(2a)#

#cos x=(-1+-sqrt(1^2-4*(2)*(-1)))/(2*2)#

#cos x=(-1+-3)/4#

#cos x_1=1/2# and #cos x_2=-1#

#x_1=60^@, 300^@#
#x_2=180^@#

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