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Can you solve this trigonometric equation: tan2x = 8cos^2x - cotx?The intervals are [0,...

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nerdynikki | Student, Undergraduate | eNoter

Posted November 30, 2011 at 5:51 AM via web

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Can you solve this trigonometric equation: tan2x = 8cos^2x - cotx?

The intervals are [0, pi/2]

If you can solve this, you're my hero! I've been working on this for hours and I still can't get it.

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sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted December 1, 2011 at 12:46 AM (Answer #1)

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Write `tan 2x = tan (x+x) = (tan x +tan x)/(1 - tan^2 x) = (2 tan x)/(1 - tan^2 x)`

Use the formula `cos^2 x = 1/(1+ tan^2 x)`

Write `cot x = 1/tan x`

Use a letter to substitute tan x.

tan x = y

Write the equation again and use y instead of tan x.

`(2y)/(1-y^2) = 8/(1+y^2) - 1/y`

`2y*y*(1+y^2) = 8y(1-y^2) - (1-y^2)(1+y^2)`

Use the difference of squares to write (`1-y^2` )(`1+y^2` ) = `1 - y^4`

Open the brackets:

`2y^2 + 2y^4 = 8y - 8y^3 - 1 + y^4`

`` `y^4- 8y^3 + 2y^2 - 8y + 1 = 0`

Write `2y^2 = y^2 + y^2`

(`y^4- 8y^3 + y^2` ) + (`y^2- 8y + 1` ) = 0

`y^2` *(`y^2- 8y + 1` ) + (`y^2- 8y + 1` ) = 0

(`y^2- 8y + 1` )*(`y^2 + 1` ) = 0 => `y^2- 8y + 1 = 0 or y^2 + 1`  = 0

Since you need to find real roots, you will not consider the equation `y^2 + 1 = 0`  that yields complex roots.

Solve the equation `y^2- 8y + 1 = 0` .

`y_(1,2) = [8+-sqrt(64-4)]/2`

`` `y_(1,2) = [8+-2sqrt(15)]/2`

`y_(1,2) = 4+-sqrt(15)`

tan x = `4+sqrt(15)`  => x = arctan  `(4+sqrt(15))`  = 82.76`^o`

tan x = `4-sqrt(15)`  => x = arctan `(4-sqrt(15)) ` = 7.23`^o`

The solutions of the equation are: {7.23`^o`  ; 82.76`^o` }

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