Question

Please help me find all solutions in the interval [0, 2π) for these equations:

1) sin^2 x cos x = 4 cosx


2) 1 - cos x = √3 sinx

I need help.

Answer 1

1) the obvious is when cosx=0, or x=90,270 deg
then for other angles,
sin^2x=4
sinx=+-2 which has no solutions.

2) 1- cosx = sqrt3 sinx
sqrt3 sinx + cosx =1
We may use the concept of subsidary angles.
Let sqrt3sinx + cosx = rsin(x + a)
sqrt3sinx + cosx = rsinxcosa+rsinacosx
Comparing the coefficient,
rcosa = sqrt3...(1)
rsina = 1...(2)
(1)^2+(2)^2
r^2=4
r=2(take it positive, you may take negative too)
cosa = sqrt3 /2
sina = 1/2
Obviously a=30
Therefore
sqrt3sinx + cosx = 2sin(x + 30)

sqrt3 sinx + cosx =1
2sin(x + 30)=1
sin(x+30)=1/2
x+30 = 180n + (-1)^n (30)
x = 180 n + (-1)^n (30) - 30