Solve the equation #cos^2x-sin^2x=sinx+1# in the interval #(0<=x<=2pi)#?

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2 Answers
Shwetank Mauria
Oct 4, 2017

#x={0,(5pi)/6,pi,(7pi)/6,2pi}#

Explanation:

As #cos^2x-sin^2x=sinx+1#

we have #1-sin^2x-sin^2x=sinx+1#

or #2sin^2x+sinx=0#

or #sinx(2sinx+1)=0#

i.e. either #sinx=0=sin0# i.e. #x=npi#

or #sinx=-1/2=sin(-pi/6)# i.e. #x=npi+(-1)^n(-pi/6)#

and with in period #[0,2pi]#, we have

#x={0,(5pi)/6,pi,(7pi)/6,2pi}#

Jacq
Oct 4, 2017

#x=0, (5pi)/6, pi, (7pi)/6, 2pi# #(0<=x<=2pi)#

Explanation:

#cos^2x-sin^2x=sinx+1#
#(1-sin^2x)-sin^2x=sinx+1#
#-2sin^2x cancel(+1)=sinx cancel(+1)#
#2sin^2x+sinx=0#
#sinx(2sinx+1)=0#

Here we the above equation to get two new equations (I'm assuming you know how to solve quadratic equations):

#sinx=0#
#therefore x=0, pi, 2pi# #(0<=x<=2pi)#

OR

#2sinx+1=0#
#sinx=-1/2#

Here, we cannot go straight to saying #x=pi/6#, as this would give a positive answer. Sine is negative in the 3rd and 4th quadrants, so these of #x# in #sinx=-1/2# can be found by putting the acute positive angle #x=pi/6# into #pi+x#, and #2pi-x#.

So,

#2sinx+1=0#
#sinx=-1/2#
#x= pi+pi/6, 2pi-pi/6#
#therefore x=(5pi)/6, (7pi)/6# #(0<=x<=2pi)#

Putting these two sets of answers together:
#therefore x=0, (5pi)/6, pi, (7pi)/6, 2pi# #(0<=x<=2pi)#

However this is assuming that the domain the question asked for is inclusive of #0# and #2pi#.

If it is exclusive of these, the solutions to this equation will be:
# x= (5pi)/6, pi, (7pi)/6# #(0 < x < 2pi)#

Solving this is just like solving an algebraic equation once you get everything in terms of #sinx#, so be sure to know your algebra and trig identities back to front!

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