How do I solve 2secxtanx+2secx+tanx+1=0
- print Print
- list Cite
Expert Answers
calendarEducator since 2011
write5,349 answers
starTop subjects are Math, Science, and Business
You need to group the terms such that:
`(2secxtanx+2secx)+(tan x + 1) = 0 `
You need to factor out `2sec x` in the first group such that:
`2sec x(tan x +1) + (tan x + 1) = 0 `
You need to factor out `tan x + 1` such that:
`(tan x + 1)(2sec x + 1) = 0`
You need to consider the first factor equal to zero such that:
`tan x +1 = 0 => tan x = -1 => x = arctan(-1) + npi`
Using the trigonometric identity `arctan(-alpha) = -arctan alpha` yields:
`x = -arctan 1 + npi => x = -pi/4 + npi`
You need to consider the second factor equal to zero such that:
`2sec x + 1 = 0 => 2sec x = -1 => sec x = -1/2`
Since `sec x = 1/cos x => 1/cos x = -1/2 => cos x = -2`
The equation `cos x = -2` is invalid since the values of cos x cannot be smaller than -1.
Hence, evaluating the general solutions to the given trigonometric equation yields `x = -pi/4 + npi` .
Related Questions
- How do I solve sinh(x)=1
- 1 Educator Answer
- Solve cosx tanx=1/2
- 1 Educator Answer
- Prove the identity: `tanx/(secx+1) = (secx-1)/tanx`
- 1 Educator Answer
- Verify: (cosx/1 + sinx) + (cosx/1-sinx) = 2secx
- 1 Educator Answer
- How do I solve 6 3/4 + 1 5/6?
- 2 Educator Answers
2secx tanx + 2secx + tanx + 1 = 0
tanx(2secx + 1) + 2secx+1 = 0
(2secx + 1)(tanx + 1) = 0
2secx + 1 = 0 or tanx + 1 = 0
secx = -1/2 or tanx = -1
tanx = -1 (ignoring secx = -1/2 which is not a valid value)
hope this will help you.../ just do pratice ... ... ... ... .. (:
Student Answers