Solve #cos^4(a)-sin^2(a)=1# ?

2 Answers
Douglas K.
Feb 8, 2018

#a = npi, n in ZZ#

Explanation:

Given: #cos^4(a)-sin^2(a)=1#

Substitute #sin^2(a) = 1-cos^2(a)#

#cos^4(a)-(1-cos^2(a))=1#

Rearrange into standard quadratic form:

#cos^4(a)+cos^2(a)-2=0#

Factor:

#(cos^2(a) -1)(cos^2(a)+2)=0#

There are two roots

#cos^2(a) =1# and #cos^2(a) = -2#

We must discard #cos^2(a) = -2#, because it implies imaginary values for the cosine function and imaginary values are not with the cosine functions range.

Use the square root on #cos^2(a) =1#

#cos(a) = +-1#

Separate into two roots:

#cos(a) = 1# and #cos(a)=-1#

Use the inverse cosine function on both sides:

#a = cos^-1(1)# and #a = cos^-1(-1)#

The above simplifies to integer multiples of #pi#:

#a = npi, n in ZZ#

Cesareo R.
Feb 8, 2018

See below.

Explanation:

#cos^2a = 1- sin^2 a# and

#(1-sin^2a)^2-sin^2a = 1-2sin^2a+sin^4a-sin^2a = 1# or

#sin^2a(sin^2a-3)=0#

Here #sin^2a -3 ne 0# hence

#sin^2a = 0 rArr a = {2k pi} uu {pi+2k pi}# for #k in ZZ#

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