How do you solve #cos^2 x-sin^2 x=sin x#?

2 Answers
Nghi N.
May 21, 2016

#(pi/6),(5pi)/6, (3pi)/2#

Explanation:

Replace in the equation #cos^2 x# by #(1 - sin^2 x)# -->
#1 - sin^2 x - sin^2 x - sin x = 0#
#-2sin^2 x - sin x + 1 = 0#
Solve this quadratic equation in sin x.
Since a - b + c = 0, use shortcut. The 2 real roots are: sin x = -1 and #sin x = -c/a = 1/2.#
a. #sin x = -1 --> x = 3pi/2#
b. #sin x = 1/2# --> 2 solution arcs -->
#x = pi/6#
#x = pi - pi/6 = (5pi)/6#
General answers:
#x = pi/6 + 2kpi#
#x = (5pi/6) + 2kpi#
#x = (3pi)/2 + 2kpi#
Checking these answers by calculator is advised.

P dilip_k
May 21, 2016

#pi/6(4k+1),pi/2(4k-1)#

Explanation:

Putting #cos^2x-sin^2x=cos2x# we have

#cos^2x-sin^2x=sinx#

#=>cos2x=sinx#
#=>cos2x=cos(pi/2-x)#

So #2x=2kpi+-(pi/2-x)#

where #k epsilon Z#

when
#2x=2kpi+(pi/2-x)#
#=>2x+x=2kpi+pi/2=pi/2(4k+1)#
#=>3x=pi/2(4k+1)#
#=>x=pi/6(4k+1)#

Again when
#2x=2kpi-(pi/2-x)#
#=>2x=2kpi-pi/2+x#
#=>x=2kpi-pi/2=pi/2(4k-1)#

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