Trigonometry Question

Question

4) The solution to 1+ sin x = cos x, can be written in the form a(pi)n, n E Z, b/c(pi) + d(pi)m, m E Z, find the values of a, b, c, and d.

Answer 1

1 + sinx = cosx

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

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

1 + sinx = sqr(1-2sin^2x) square both sides, but be aware you may be introducing an extraneous solution which is not a solution to the original equation. that happens when you multiply by a variable.

1 + 2sinx + sin^2x = 1 - 2sin^2x

3sin^2x +2sinx = 0 factor out sinx

(sinx)(3sinx+2) = 0 set each factor =0

sinx = 0, x = 0 or 2npi where n= any integer, while pi looks like a solution, it isn't it, if you try to plug it into the original equation it doesn't work.

x=3pi/2 does work. 1 - sin(3pi/2) = cos(3pi/2)

1- 1 = 0

3sinx =-2

sinx =-2/3

x = a quadrant III or IV angle where cosx = + or - sqr5/3

try that in the original equation

1 - (-2/3) = (1/3)sqr5

5/3 =sqr5/3

it doesn't work

only solutions are 2npi and 3pi/2 + 2npi = (3/2 + 2n)pi where n = any integer

apin = 2pin where a= 2

(b/c)pi + d(pi)n = (3/2)pi + 2(pi)n where b=3, c=2 and d = 2