math.answers.com/math-and-arithmetic/How_do_you_solve_6tanx2-17tanx_plus_7_equals_0

Preview meta tags from the math.answers.com website.

Linked Hostnames

8

Thumbnail

Search Engine Appearance

Google

https://math.answers.com/math-and-arithmetic/How_do_you_solve_6tanx2-17tanx_plus_7_equals_0

How do you solve 6tanx2-17tanx plus 7 equals 0? - Answers

Assuming your equation is: 6(tanx)2-17tanx + 7 = 0The easiest way to do it is use a substitution. Let s=tanx, then substitute s for tanx in the original equation to get the following:6s2-17s+7=0Using the quadratic equation solve for s:s = {-(-17) +- SQRT [(-17)2 - 4*(6)*7)]}/(2*6)s = [17 + SQRT (121)]/12 & s= [17 - SQRT(121)]/12s = 28/12 or 7/3 & s = 6/12 or 1/2Now substitute tanx back for s to gettan x = 7/3 & tan x = 1/2x = tan-1(7/3) or approx. 66.8o & x= tan-1(1/2) or approx. 26.6oIn radians it would be 1.166 & 0.4636



Bing

How do you solve 6tanx2-17tanx plus 7 equals 0? - Answers

https://math.answers.com/math-and-arithmetic/How_do_you_solve_6tanx2-17tanx_plus_7_equals_0

Assuming your equation is: 6(tanx)2-17tanx + 7 = 0The easiest way to do it is use a substitution. Let s=tanx, then substitute s for tanx in the original equation to get the following:6s2-17s+7=0Using the quadratic equation solve for s:s = {-(-17) +- SQRT [(-17)2 - 4*(6)*7)]}/(2*6)s = [17 + SQRT (121)]/12 & s= [17 - SQRT(121)]/12s = 28/12 or 7/3 & s = 6/12 or 1/2Now substitute tanx back for s to gettan x = 7/3 & tan x = 1/2x = tan-1(7/3) or approx. 66.8o & x= tan-1(1/2) or approx. 26.6oIn radians it would be 1.166 & 0.4636



DuckDuckGo

https://math.answers.com/math-and-arithmetic/How_do_you_solve_6tanx2-17tanx_plus_7_equals_0

How do you solve 6tanx2-17tanx plus 7 equals 0? - Answers

Assuming your equation is: 6(tanx)2-17tanx + 7 = 0The easiest way to do it is use a substitution. Let s=tanx, then substitute s for tanx in the original equation to get the following:6s2-17s+7=0Using the quadratic equation solve for s:s = {-(-17) +- SQRT [(-17)2 - 4*(6)*7)]}/(2*6)s = [17 + SQRT (121)]/12 & s= [17 - SQRT(121)]/12s = 28/12 or 7/3 & s = 6/12 or 1/2Now substitute tanx back for s to gettan x = 7/3 & tan x = 1/2x = tan-1(7/3) or approx. 66.8o & x= tan-1(1/2) or approx. 26.6oIn radians it would be 1.166 & 0.4636

  • General Meta Tags

    22
    • title
      How do you solve 6tanx2-17tanx plus 7 equals 0? - Answers
    • charset
      utf-8
    • Content-Type
      text/html; charset=utf-8
    • viewport
      minimum-scale=1, initial-scale=1, width=device-width, shrink-to-fit=no
    • X-UA-Compatible
      IE=edge,chrome=1

  • Open Graph Meta Tags

    7
    • og:image
      https://st.answers.com/html_test_assets/Answers_Blue.jpeg
    • og:image:width
      900
    • og:image:height
      900
    • og:site_name
      Answers
    • og:description
      Assuming your equation is: 6(tanx)2-17tanx + 7 = 0The easiest way to do it is use a substitution. Let s=tanx, then substitute s for tanx in the original equation to get the following:6s2-17s+7=0Using the quadratic equation solve for s:s = {-(-17) +- SQRT [(-17)2 - 4*(6)*7)]}/(2*6)s = [17 + SQRT (121)]/12 & s= [17 - SQRT(121)]/12s = 28/12 or 7/3 & s = 6/12 or 1/2Now substitute tanx back for s to gettan x = 7/3 & tan x = 1/2x = tan-1(7/3) or approx. 66.8o & x= tan-1(1/2) or approx. 26.6oIn radians it would be 1.166 & 0.4636

  • Twitter Meta Tags

    1
    • twitter:card
      summary_large_image

  • Link Tags

    16
    • alternate
      https://www.answers.com/feed.rss
    • apple-touch-icon
      /icons/180x180.png
    • canonical
      https://math.answers.com/math-and-arithmetic/How_do_you_solve_6tanx2-17tanx_plus_7_equals_0
    • icon
      /favicon.svg
    • icon
      /icons/16x16.png

Links

59