How do you solve $2 sec x tan x + 2 sec x + tan x + 1 = 0$ $2secxtanx+2secx+tanx+1=0$ in the interval [0,360]?

Question

4306 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-09-08T13:38:17

The solution set is ${\frac{3 π}{4}, \frac{7 π}{4}}$ ${(3pi)/4, (7pi)/4}$ .

Factor:

$2 sec x (tan x + 1) + 1 (tan x + 1) = 0$ $2secx(tanx + 1) + 1(tanx + 1) = 0$

$(2 sec x + 1) (tan x + 1) = 0$ $(2secx + 1)(tanx + 1)= 0$

$sec x = - \frac{1}{2} and tan x = - 1$ $secx = -1/2 and tanx = -1$

$cos x = - 2 and tan x = - 1$ $cosx = -2 and tanx = -1$

$x = \emptyset and x = \frac{3 π}{4} and \frac{7 π}{4}$ $x = O/ and x = (3pi)/4 and (7pi)/4$

Hence, the solution set is ${\frac{3 π}{4}, \frac{7 π}{4}}$ ${(3pi)/4, (7pi)/4}$ .

Hopefully this helps!

How do you solve 2secxtanx+2secx+tanx+1=02secxtanx+2secx+tanx+1=02secxtanx+2secx+tanx+1=0 in the interval [0,360]?