6 cos2x + (2k-1) sin 2x = 8 k?
- print Print
- list Cite
Expert Answers
calendarEducator since 2011
write5,349 answers
starTop subjects are Math, Science, and Business
You should use the following formulas such that:
`sin 2x = (2tan x)/(1 + tan^2 x)`
`cos 2x= (1 - tan^2 x)/(1 + tan^2 x)`
You need to substitute `t` for `tan x` such that:
`6*(1 - t^2)/(1 + t^2) + (2k-1)*(2t)/(1 + t^2)= 8`
`6 - 6t^2 + 4kt - 2t - 8 - 8t^2 = 0`
`-14t^2 + 2t*(2k-1) - 2 = 0`
Dividing by -2 yields:
`7t^2 - t(2k-1) + 1 = 0`
You should use quadratic formula to evaluate t such that:
`t_(1,2) = (2k-1 +- sqrt((2k-1)^2 - 28))/14`
You should notice that the radicand`(2k-1)^2 - 28` needs to be positive for`t_1` and `t_2` to be real values such that:
`(2k-1)^2 - 28 >= 0`
Solving the attached equation yields:
`(2k-1)^2 - 28 = 0 => (2k-1)^2 = 28 => 2k-1 = +-sqrt28`
`2k-1 = +-2sqrt7 => 2k = 1+-2sqrt7 => k = (1+-2sqrt7)/2`
Notice that the expression is positive for `k in (-oo, (1-2sqrt7)/2)U((1+2sqrt7)/2, +oo).`
Hence, there exists solution to the given trigonometric equation if `k in (-oo, (1-2sqrt7)/2)U((1+2sqrt7)/2, +oo).`
Related Questions
- How to prove the identity `sin^2x + cos^2x = 1` ?
- 3 Educator Answers
- Evaluate the integral of function y=cos2x/cos^2x*sin^2x.
- 1 Educator Answer
- Find sin 2x, cos 2x, and tan 2x from the given information.Find sin 2x, cos 2x, and tan 2x from...
- 1 Educator Answer
- write sin2x and cos2x w.r.t. tangent? calculate 2^1/2 sin2x+3^1/2 cos2x, tanx=6^1/2/3
- 1 Educator Answer
- Verify: tan^2x - sin^2x= (tan^2x)(sin^2x)
- 1 Educator Answer
Unlock This Answer Now
Start your 48-hour free trial to unlock this answer and thousands more. Enjoy eNotes ad-free and cancel anytime.