Pls can someone find all the values of x solve for 0° < x < 360°... 2 sin^2 x = cos x +1 ?

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Answer 1 · 2017-12-31T02:17:58

#x=60^@, 180^@, and 300^@#

.

#2sin^2x=cosx+1#

We have an identity that says:

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

We solve for #sin^2x#:

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

We substitute this n the problem equation:

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

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

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

Using the quadratic formula, we solve for #cosx#:

Our equation is in the form:

#ax^2+bx+c=0#

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

In our problem, #a=2#, #b=1#, and #c=-1#

#cosx=(-1+-sqrt(1-4(2)(-1)))/4=(-1+-sqrt9)/4=(-1+-3)/4#

#cosx=-1# which means #x=180^@#

#cosx=1/2# which means #x=60^@, 300^@#