solve: cosx + tanx = 5√3/6 for 0 ≤ x ≤ 2π
3 answers
x = 2π/3
But that answer does not check out in the equation,
can you show me the steps.
btw, the right side of my equation is (5√3) / 6
can you show me the steps.
btw, the right side of my equation is (5√3) / 6
cosx + tanx = 5√3/6
cos^2x + 2sinx + tan^2x = 25/12
Now let u = cosx and you have
u^2 + 2√(1-u^2) + 1/u^2 = 37/12
u = ±√3/2
so x = π/6, 5π/6, 7π/6, 11π/6
since we squared twice, we may have some extraneous solutions, so check them.
cos π/6 + tan π/6 = √3/2 + 1/√3 = 5√3/6
There is also a solution at x=4.23 ≈ 4π/3
had to do some approximation there, but I knew there had to be one in QIII, at some place where tanx is big enough to overcome the minus value of cosx
cos^2x + 2sinx + tan^2x = 25/12
Now let u = cosx and you have
u^2 + 2√(1-u^2) + 1/u^2 = 37/12
u = ±√3/2
so x = π/6, 5π/6, 7π/6, 11π/6
since we squared twice, we may have some extraneous solutions, so check them.
cos π/6 + tan π/6 = √3/2 + 1/√3 = 5√3/6
There is also a solution at x=4.23 ≈ 4π/3
had to do some approximation there, but I knew there had to be one in QIII, at some place where tanx is big enough to overcome the minus value of cosx