# Thread: qn done with different methods obtaining diff ans

1. ## qn done with different methods obtaining diff ans

i worked it out on a paper. pls see the pic attached.thanks

2. (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.