Question

solve sec^2x+5tanx=-2

Answer 1

sin^2(x)+cos^2(x)=1
so
tan^2(x)+1=sec^2(x)
so
we have
tan^2(x)+1+5tanx=-2
tan^2(x)+5tanx+1+2=0
(tanx)^2+5tanx+3=0
let u=tanx
so u^2=tan^2(x) or (tanx)^2
u^2+5u+3=0
solve this for u first

Answer 2

1+tan^2x+5tanx=-2
tan^2x+5tanx+3=0
tanx=(-5+-sqrt(25-12))/2

Answer 3

hello got it??

Answer 4

is that the final answer?

Answer 5

you want to solve for x
take tan inverse of both sides

Answer 6

so (tan^-1x)^2 +5tan^-1x+3?

Answer 7

no

Answer 8

im confused...

Answer 9

tanx=(-5+-sqrt(25-12))/2
take tan inverse (or arctan) of both sides

Answer 10

u have found the tanx=(-5+-sqrt(25-12))/2
then x=tan^{-1}((-5+-sqrt(25-12))/2)

Answer 11

thanks for the help myininaya and dipankarstudy

Answer 12

good luck...