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How do you solve #1 - 2(sinx)^2 = cosx, 0 <= x <= 360#. Solve for #x#?

Answer 1

Madelyn Adams

#x=0,120,240,360#

#asin^2x+acos^2x-=a#

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

#1-(2-2cos^2x)=cosx#

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

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

Substitute #u=cosx#

#2u^2-u-1=0#

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

#u=(1+-sqrt(1-4(-2)))/4#

#u=(1+-sqrt(1+8))/4#

#u=(1+-sqrt(9))/4#

#u=(1+-3)/4#

#u=1or-1/2#

#cosx=1or-1/2#

#x=cos^-1(1)=0,(360-0)=0,360# #x=cos^-1(-1/2)=120,(360-120)=120,240#

#x=0,120,240,360#

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Answer 2

Charles Autterson

#color(blue)(0, 120^@,240^@,360^@)#

Identity:

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

Substituting # (1-cos^2x)# in given equation:

#1-2(1-cos^2x)=cosx#

Subtracting #cosx# and expanding:

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

Simplify:

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

Let # \ \ \u=cosx#

#:.#

#2u^2-u-1=0#

Factor:

#(2u+1)(u-1)=0=>u=-1/2 and u=1#

But #u=cosx#

#:.#

#cosx=-1/2 , cosx=1#

#x=arccos(cosx)=arccos(-1/2)=>x=120^@#

This is in quadrant II, we also have an angle in quadrant III:

#360^@-120^@=240^@#

#x=arccos(cosx)=arccos(1)=>x=0, 360^@#

Collecting solutions:

#color(blue)(0, 120^@,240^@,360^@)#

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

How do you solve #1 - 2(sinx)^2 = cosx, 0 &lt;= x &lt;= 360#. Solve for #x#?

How do you solve #1 - 2(sinx)^2 = cosx, 0 <= x <= 360#. Solve for #x#?