(i) show that 5+3tan^2 x/1+tan^2x=3+2cos^2x. (ii) hence solve the equation 5+3 tan^2 x/1+tan^2 x=7cos x,for 0 is less than equals to x is less than equals to 360.

Answer 1

i)

(5+3tan²x)/(1+tan²x) = 3+2cos²x

(2+3 + 3tan²x)/sec²x

(2+3(1+tan²x))/sec²x

(2+3sec²x)/sec²x

2cos²x+3 = 3+2cos²x

ii)

(5+3tan²x)/(1+tan²x) = 7cosx

2cos²x+3 = 7cosx

2cos²x - 7cosx + 3 = 0

Which is a quadratic equation

cosx = (7±√(49-24))/4 = 13/4, 1/2

13/4 not in domain for cosine, so can be ignored

For cosx = 1/2:

x = 60°, 300°