How do you solve sin^2 = 2cosx - 2 algebraically?

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Nov 23, 2017

Answer:

#x=color(blue)(2npi)# ; where #color(red)(n in ZZ)# , the set of integers.

Explanation:

I am assuming the question is :

#sin^2x=2cosx-2#

We know the trigonometric identity#rarr#

#implies color(red)(sin^2x+cos^2x=1)#

Subtract #cos^2x# from both sides to get :

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

Put this value of #sin^2x# in the given equation :

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

Add #cos^2x# both sides , we get :

#implies 1-cancel(cos^2x)+cancel(cos^2x) = cos^2x+2cosx-2#

Now subtract #1# from both sides :

#implies cancel(1)-cancel(1)=cos^2x+2cosx-2-1#

Now we have a quadratic equation in #color(blue)(cos^2x) rarr#

#implies cos^2x+2cosx-3=0 #

We solve it by factorisation method (splitting the middle term) :

#implies cos^2x+3cosx-cosx -3=0#

#implies cosx(cosx+3) -1(cosx+3)=0#

#implies (cosx-1)(cosx+3)=0#

This gives us two solutions :

  • #cosx=-3# (rejected) , because #color(blue)(cosx in [-1,+1]# .
  • #cosx= 1# (accepted)

The second one is the solution ,

#therefore color(red)(cosx=1)# , which has the general solution#rarr#

#implies color(blue)(x=2npi)# , where #color(blue)(n in ZZ)# .

Moreover you can solve it graphically also #rarr#

Here are the curves of #color(red)(y=sin^2x)# and #color(blue)(y=2cosx-2) rarr#

Note : The colors of the graphs have been mentioned.

Meta calculator

Fun Fact : Here #"x-axis" or y=0# is the common tangent to the
curves wherever they touch , i.e #(forall x=2npi , n in ZZ)# .

For further references , read this : solving trigonometric equations

Hope it helps #: )#

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