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Find the exact solution to the equation in the interval [0,2pi)4sinxcosx=1

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rc1994 | Student, College Freshman | Honors

Posted January 4, 2012 at 8:27 AM via web

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Find the exact solution to the equation in the interval [0,2pi)

4sinxcosx=1

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embizze | High School Teacher | (Level 1) Educator Emeritus

Posted January 4, 2012 at 1:46 PM (Answer #1)

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Given `4sinxcosx=1` on `[0,2pi]`

(1) Since `sin2x=2sinxcosx` we can rewrite as `2sin2x=1`

Then `sin2x=1/2`

(2) The period for `sin2x` is `pi` , so we find solutions to `sin^(-1)(1/2)` on `[0,4pi]` , so `2x=pi/6,(5pi)/6,(13pi)/6,(17pi)/6` or `x=pi/12,(5pi)/12,(13pi)/12,(17pi)/12`

So the solution is `x=pi/12,(5pi)/12,(13pi)/12,(17pi)/12`

The graphs of 4sinxcosx and y=1:

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hala718 | High School Teacher | (Level 1) Educator Emeritus

Posted January 4, 2012 at 8:33 AM (Answer #2)

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Given the trigonometric equation:

`4sinxcosx = 1`

`` We will rewrite :

`2*2sinxcosx = 1`

But we know that `sin2x = 2sinxcosx`

`==gt 2sin2x = 1`

Now we will divide by 2.

`==gt sin2x = 1/2 `

`==gt 2x = pi/6 , (5pi)/6` ( sin is positive in 1st and 2nd quadrants)

Now we will divide by 2.

`==gt x = pi/12 , (5pi)/12`

``

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