solving trig equation?

Question

3^1/2(tan^2 x) + (2+5(3^1/2))sec x + 10 + 3^1/2= 0

3^1/2 = square root of 3

Right now I have tan^2 x + 2/[{3^1/2} cosx] + 5/cos x = (-10(3^1/2))/3

I was going to sub tan^2 x with (1-cos 2x)/ (1+cos 2x) but it seems like I'm just going to end up going in circles?

Answer 1

3^1/2*tan^2(x) + (2+5(3^1/2))*sec(x) + 10 + 3^1/2= 0

Express the tan(x) as sin(x)/cos(x) and the sec(x) as 1/cos(x) and then put everything over a common denominator of cos^2(x):

0 = [3^1/2*sin^2(x) + (2 + 5*3^1/2)*cos(x) +10*cos^2(x) + 3^1/2*cos^2(x)]/cos^2(x)

Note that 3^1/2*sin^2(x) +3^1/2*cos^2(x) = 3^1/2 * 1

0 = [10*cos^2(x) + (2 + 5*3^1/2)*cos(x) + 3^1/2]/cos^2(x)

multiply through by cos^2(x) (we need to watch out, however, for situations when cos^2(x) = 0)

0 = [10*cos^2(x) + (2 + 5*3^1/2)*cos(x) + 3^1/2]

Let u = cos(x), then we get the quadratic algebraic equation:

0 = 10u^2 + (2 + 5*3^1/2)u + 3^1/2

Use the quadratic formula to find that the roots of this equation. The roots of the original equation are given by the arccos of the values of u you found from solving the quadratic.

Answer 2

3^1/2(tan^2 x) + (2+5(3^1/2))sec x + 10 + 3^1/2= 0

using: tan^2 x=Sec^2 x-1

3^1/2 ( Sec^2 - 1) + Sec ( 2+ 5(3^1/2) + 10 + 3^1/2 = 0
expanding and let 3^1/2 = z

z Sec^2 - z + 2 Sec +5 z Sec + 10 + z = 0 Cancel z - z

z Sec^2 + 2Sec + 5z Sec + 10 = 0

Sec( Sec z + 2 + 5z) = -10

Answer 3

Did u try substituting

tan^2 x=Sec^2 x-1 at the beginning....

so as to have only sec x as in a quadratic eqn..

Answer 4

3^1/2(sec^2x-1)+2+5(3)^1/2secx+3^1/2=0
dividing throughout by (3)^1/2
(sec^2x-1)+5secx+1+2(3)^-1/2=0
sec^2x+5secx+2(3)^-1/2=0
secx=[-5+/-(25-8(3)^-1/2]/2
now using a calculator you can find the value of secx