anonymous
  • anonymous
Solve the following Trig equation. -3-3sin(theta)=cos^2(theta)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
Re-write cos^2 in terms of sin^2 and you have a quadratic!
anonymous
  • anonymous
so i need to solve the LHS?
anonymous
  • anonymous
No. Use the fact cos^2x = 1 - sin^2x, and then move it all to one side and solve as a quadratic

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anonymous
  • anonymous
Im not sure how to do that?
anonymous
  • anonymous
I'm using x as theta. -3-3sinx = cos^2x => -3-3sinx = 1-sin^2x =>sin^2x - 3sin - 4 = 0
anonymous
  • anonymous
Ahh i see how you got it now, now how do i set it up to find on all the solutions from 0
anonymous
  • anonymous
Just solve it as you would the quadratic y^2 - 3y -4, for sin x. Note you need solutions -1 <= sin(x) <= 1 , you can ignore the others. arcsin this gives you the first solution Generally, if you find one solutions 0
anonymous
  • anonymous
i have to show my work in radians, how do i get the value to use for 2Kpi
anonymous
  • anonymous
wut? sin^2x -3sinx - 4 = (sin(x) - 4)(sin(x) + 1) So sin x = ... or ... However -1 =< sin x =< 1 , so we can ignore one of these. And then use sin^-1 to work out the values of x.
anonymous
  • anonymous
My questtion says solve each of the following trignometric questions on interval 0
anonymous
  • anonymous
Is the problem that you normally work in degrees, or do you just not know how to find the angle?
anonymous
  • anonymous
i just dont know how to find the angle on this type of problem
anonymous
  • anonymous
Oh, OK. Well can you see what the value of sin(x) is from above?
anonymous
  • anonymous
... I've got to go, so I'll rush this, sorry. From the factorisation, sin(x) = -1 (we can ignore sin(x) = 4 because it's too big) We use the inverse sin / arcsin / sin^-1 function => arcsin(-1) = -pi/2 But we need a value between 0 and 2pi 20 we can add 2pi => 3/2 pi this is the only value in the inteval.
anonymous
  • anonymous
Okay thank you for your time!

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