Question

Solve the following equations for x between 0 and 360 degrees
(a) sinx + 4cos^2x = 1
(b) 2sec^2x = 3 - tan^2x
For part (a) I have gotten this far: (4sinx + 3) (sinx - 1) = 0 How to I solve for x from there?
Same applies to part (b): (tanx - 1) (tanx + 3) = 0. How to solve for x? Please show all working thanks.

Answer 1

you said, sin(x) + 4(sin^2(x) - 1) = 1
isn't it supposed to be
sin^2(x) + cos^2(x) = 1
so if you're gonna replace cos^2(x), it will be 1- Sin^2(x) not sin^2(x) - 1
that will be -cos^2(x) ?

Answer 2

sin(x) + 4(1 - sin^2(x)) = 1
sin(x) + 4 - 4sin^2(x) = 1
sin(x) +3 - 4sin^2(x) = 0
4 sin^2(x) - sin(x) - 3 = 0
4 sin(x)^2 - sin(x) - 3 = 0
4x^2 - x - 3 = 0
4x^2 - 4x + 3x - 3 = 0
4x(x - 1) + 3(x - 1) = 0
(x - 1)(4x + 3) = 0

x - 1 = 0
4x + 3 = 0

sin(x) - 1 = 0
4(sin(x)) + 3 = 0

sin(x) = 1
sin(x) = -3/4

x = sin^{-1}(1)
x = sin^{-1}(-3/4)

x = pi/2
x = no solution

Answer 3

@ambitiouschick