i worked it out on a paper. pls see the pic attached.thanks
(1 + sin x)^2 + cos^2 x = 4(1 + sin x)(cos x)
a simple substitution will confirm that this one is not an identity, like x = 90 degrees will give, 4 + 0 = 4(1 + 1)(0) is not true. So, this is a trigonometric equation, . . . .
expand the LHS of the equation,
LHS = 1 + 2sin x + sin^2 x + cos^2 x
= 1 + 2sin x + sin^2 x + cos^2 x
= 1 + 2sin x + 1
= 2 + 2sin x
= 2(1 + sin x),
then 2(1 + sin x) = 4(1 + sin x)(cos x),
2(1 + sin x) - 4(1 + sin x)(cos x)= 0
2(1 + sin x)(1 - 2cos x) = 0,
for the first factor sin x = -1,
x = 270 degrees
for the factor 1 - 2cos x = 0,
x = 60 degrees and 300 degress
Solution Set : {60, 270, 300}
i speculate that you introduced an extraneous result in your solution 1, by introducing this identity, sec x = 1/cos x. thus that cos THING appeared, whose value is 90 degrees.
See my graph, that wil confirm your second solution as correct. . . . the curve intersects the x-axis just 3 times from 0 to 2pi interval.