Inverse trig.

  • Thread starter zaddyzad
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  • #1
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Homework Statement



I have two questions sin2x = 1/25 and this obviously becomes sinx= +-(1/5)
I also have cos2-1.5cosx-0.54 and cosx = (-3/10) and (9/5)

Now this is asking for me to solve for the x value in radians in the domain [0,2pi] and I have no idea how to solve these for exact values. Help would be appreciated.

Homework Statement

 

Answers and Replies

  • #2
34,687
6,394

Homework Statement



I have two questions sin2x = 1/25 and this obviously becomes sinx= +-(1/5)
I also have cos2-1.5cosx-0.54 and cosx = (-3/10) and (9/5)
What does "cos2-1.5cosx-0.54 " mean?
cos(x) can't possibly equal 9/5.
Now this is asking for me to solve for the x value in radians in the domain [0,2pi] and I have no idea how to solve these for exact values. Help would be appreciated.

Homework Statement

 
  • #3
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that was my second equation, and my bad for putting it down as it is an extraneous root.
 
  • #4
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6,394
"cos2-1.5cosx-0.54" is NOT an equation.

Did you mean cos2(x) - 1.5cosx-0.54 = 0?

Are there two separate questions, or do you have a question about a system of two equations?

Help us out here - don't make us guess about this stuff...
 
  • #5
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Sorry, it is two questions. And yes those are my two problems that I need to solve over the domain [0,2pi]
I need help solving for x using inverse trig.
 
  • #6
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For the first question, if you had sin(x) = ±1/4, there are two numbers in [0, ##2\pi##] for which sin(x) = 1/4 and two more in this interval for which sin(x) = -1/4.

If we let θ be the smallest of the four values, we have sin(θ) = 1/4, so θ = sin-1(1/4). The other value in that interval for which sin(θ) = 1/4 is ##\pi - \theta##, or ##\pi - sin^{-1}(1/4)##. These are the exact values.

Similar work will get you the two values for which sin(θ) = -1/4.

Your first problem is similar to this.
 

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