What is algebraic method in solving equation 2cos2x=8cosx -1?

Asked on by luvgoj

sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted on

You need first to use the double angle formula such that:

`cos 2x = 2cos^2 x - 1`

Replacing `2cos^2 x - 1` for `cos 2x` to the left side, yields:

`2(2cos^2 x - 1) = 8cos x - 1`

You need to expand the left side, such that:

`4cos^2 x - 2 = 8cos x - 1 => 4cos^2 x - 8cos x - 2 + 1 = 0`

`4cos^2 x - 8cos x - 1 = 0`

You should convert the trigonometric equation in a quadratic algebraic equation, replacing the variable y for the function cos x, such that:

`4y^2 - 8y - 1 = 0`

Using quadratic formula, yields:

`y_(1,2) = (8 +- sqrt(64 + 16))/8 => y_(1,2) = (8 +- sqrt80)/8`

`y_(1,2) = (8 +- 4sqrt5)/8`

`y_(1,2) = (2+-sqrt5)/2`

`y_1 = (2-sqrt5)/2 ~~ 0.118`

`y_2 = 2.118`

You need to solve for x the following equations, such that:

`cos x = y_1 => cos x = 0.118 => x = +-cos^(-1)(0.118) + 2npi`

`cos x = y_2 => cos x = 2.118` invalid

Hence, evaluating the general solution to the given equation, yields `x = +-cos^(-1)(0.118) + 2npi.`

We’ve answered 319,654 questions. We can answer yours, too.
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}