Solving a difficult trig equation, help

solve the equation in the range: 0 ≤ x ≤ 360

6sec²x = 5(tanx + 1)

I have rearranged it to :

6sec²x - 5tanx = 5 , but I'm clueless as to what to do next

enchante -

Your first adjustment to get all the trig functions the same. In your problem, try to change sec x to tan x.

6sec²x - 5tanx - 5 = 0

6( 1 + tan²x) - 5 tan x - 5 = 0

6 tan²x - 5 tan x +1 = 0, now split the middle so that you can solve by grouping:

6 tan²x - 2 tan x - 3 tan x +1 = 0

2 tan x(3 tan x - 1) - (3 tan x - 1) = 0

(2 tan x - 1)(3 tan x - 1) = 0

So, solving each factor:

tan x = 1/2 or tan x = 1/3

Now just take the arctan of each and solve between 0 and 2pi

Hope that helps

Since it is quadratic, try to get it in terms of one trig ratio.
Sec^2(x) = 1+tan^2(x)

6(1+tan^2(x))-5tanx -5=0

6tan^2(x)-5tanx+1=0

(2tanx-1)(3tanx-1)=0

Tanx=1/2 or tanx = 1/3

These are calculator problems. Remember that tangent is positive in quadrants I and III.

x= 26.6, or 180+ 26.6, 18.4, 180+18.4

X= 26.6, 206.6, 18.4, 198.4

Hoping this helps!