Solve sin² θ + 3 sin θ - 1 = 0 to the nearest tenth of a degree for 0° ≤ θ ≤ 360°
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Solve sin² θ + 3 sin θ - 1 = 0 to the nearest tenth of a degree for 0° ≤ θ ≤ 360°

[From: ] [author: ] [Date: 12-05-20] [Hit: ]
3)The function y(t) = 122 sin θ - 16t² models the altitude of a ball t seconds after it was hit at the angle of θ degrees. Determine, to the nearest tenth of a degree, the measure of the angle at which the ball was hit, if it had an altitude of 40 feet after 1.5 seconds.......
Solve each equation to the nearest tenth of a degree for 0° ≤ θ ≤ 360°.

1) sin² θ + 3 sin θ - 1 = 0


2) 3 cos² θ = 2


3)The function y(t) = 122 sin θ - 16t² models the altitude of a ball t seconds after it was hit at the angle of θ degrees. Determine, to the nearest tenth of a degree, the measure of the angle at which the ball was hit, if it had an altitude of 40 feet after 1.5 seconds.

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1) sin² θ + 3 sin θ - 1 = 0
sinθ = -3 +/- sqrt( 3² - 4(1)(-1) ) / 2(1)
θ = sin^-1[( -3 +/- sqrt(13) ) / 2]
θ = 17.624° + 360°k or 162.376° + 360°k
(-3 - sqrt(13) ) / 2 is larger than 1, so for that there is no solution
answer: 17.624°, 162.376°

2) 3 cos² θ = 2
cos² θ = 2/3
cosθ = +/- sqrt(2/3)
θ = cos^-1( +/- sqrt(2/3) )
θ = +/- 35.264° + 360°k or +/- 144.736° + 360°k

3)
40 = 122sinθ - 16(1.5)²
40 = 122sinθ - 36
76 = 122sinθ
sinθ = 76/122
θ = sin^-1(76/122)
θ = 38.532° + 360°k or 141.468° + 360°k

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I can help you with the first two.

It would be good to know that the notation sin² θ is actually (sin θ)².
This in mind, I'm going to let sin θ = Y.
This is nice. From sin² θ + 3 sin θ - 1 = 0, i've now got Y² + 3Y - 1.
However, this gives horrible Y values of {-3 ± Sq Root ( 9 + 4 ) } / 2.
Although now, we can let Y = Sin θ again. Performing an inverse sin function on {-3 + Sq Root ( 9 + 4 ) } / 2 and {-3 - Sq Root ( 9 + 4 ) } / 2, we can get our angles. May I suggest a graphical calculator?

2) This is a little nicer. 3 cos² θ = 2 can be easily arranged. Moving the 3 over, gives us cos² θ = 2/3.

It may seem strange, but remembering that cos² θ is (cos θ)², we can simply sqrt 3/5. This gives us:

± sqrt [3/5]. Again, performing an inverse cos function on positive and negative sqrt [3/5] gives us our angles.

Sorry about the third question :S

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1). Let sintheta = x
Then we have x^2+3x-1=0

Use quadractic formula to get x = .3 radians.
Thus, sin theta = .3.
Arc sin produces theta = .3 radians.
Convert to degrees. .3*180/pi gives 17.6 degrees.

2). Cos^2 = 2/3. Or. Cos = sqrt 2/3 radians. Taking arccos gives .61 rads or 35.3 degrees

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2) 3cos^2x = 2
cos^2x = 2/3
cosx = √(2/3) or cosx = -√(2/3)
x = 35.3 or x = 324.7 or x = 144.7 or x = 215.3

One question at a time.
1
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