sinx = -√2/2 (x is either in quadrant 3 or quadrant 4 as sinx < 0)
x = π + π/4, 2π - π/4
= 5π/4, 7π/4
(cos2θ = sinx has an infinite variety of solutions namely x = π/2 - 2θ)
Assuming typographical mistake:
cos2x = sinx at interval [0, 2pi]
1 - 2 sin²x = sinx
So 2sin²x + sinx - 1 = 0
ie (2sinx -1)(sin x + 1) = 0
So sinx = ½, -1
so x = π/3, π - π/3, 3π/2
= π/3, 2π/3, 3π/2
cos(2θ/3) = -1 at interval [0, 2pi]
2θ/3 = π
θ = 3π/2
3cos²x - 8cosx - 3 = 0
(3cosx + 1)(cosx - 3) = 0
cosx = -1/3 (ignoring 3 as |cosx| cannot exceed one for real x)
x = π - 1.2310, π + 1.2310 (4 decimal places)
= 1.9106, 4.3726 (4 decimal places)
2006-12-14 18:17:27
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answer #1
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answered by Wal C 6
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1) sin(x) = sqrt(2)/2
Since there is no restriction to x, what you would do is solve as if there was a restriction to x in [0,2pi) and then write them as infinitely many solutions.
To solve for this question, you have to first ask yourself: on that unit circle you were asked to memorize, where is sine equal to sqrt(2)/2? The answer is in quadrants 1 and 2, and the two points are pi/3 and 2pi/3.
Therefore, you'd first write each answer.
x = {x | pi/3 .... } U {x | 2pi/3 ....}
And then you'd add 2k(pi) to represent that any number equivalent to the points on the unit circle is valid. Remember that for any point on the unit circle, you can add 2pi and have it be functionally the same for the trig functions. Therefore
x = {x | x = pi/3 + 2k(pi), for all integers k} U
{x | x = 2pi/3 + 2k(pi), for all integers k}
2) cos(2y) = sin(y), on the interval [0,2pi]
[I changed them to y instead of theta]
Your first step is to use the half angle trig identity:
cos(2x) = 1 - 2sin^2(x).
1 - 2sin^2(y) = sin(y)
Move everything over to the left hand side,
1 - 2sin^2(y) - sin(y) = 0
Write them in descending order of sine,
-2sin^2(y) - sin(y) + 1 = 0
And then multiply by -1, to make the highest power of sine positive.
2sin^2(y) + sin(y) - 1 = 0
Note that this is a quadratic in disguise, so we can factor it as a quadratic.
(2siny - 1) (siny + 1) = 0
Therefore, 2siny - 1 = 0 and siny + 1 = 0
Which solves as sin(y) = 1/2 and sin(y) = -1.
The solution set for sin(y) = 1/2 is y = {pi/6, 5pi/6}.
The solution set for sin(y) = -1 is y = 3pi/2
Therefore, y = {pi/6, 5pi/6, 3pi/2}
3) 3cos^2(x) - 8cos(x) - 3 = 0
This is a quadratic function again in disguise, and you'd have to use the quadratic formula. The difference though, is on the left hand side, you'll have cos(x) instead of just x. Therefore
cos(x) = [-b +/- sqrt(b^2 - 4ac)]/2a
And we know our coefficients.
cos(x) = [8 +/- sqrt(64 - 4(3)(-3))]/2(3)
cos(x) = [8 +/- sqrt(64 + 36)]/6
cos(x) = [8 +/- sqrt(100)]/6
cos(x) = [8 +/- 10]/6
Therefore, cos(x) = [8 +10]/6 and cos(x) = [8 - 10]/6 OR
cos(x) = 3 and cos(x) = -2/6 = -1/3
Remember that cos(x) must lie between -1 and 1, so cos(x) = 3 will have no solutions. cos(x) = -1/3 will have solutions though; they just aren't familiar on our unit circle.
Cosine is negative in quadrants 2 and 3, and your two answers will be x = cos(inverse)(-1/3) and x = pi - cos(inverse)(-1/3).
You have to plug those into your calculator, making sure it's set to RAD mode.
2006-12-14 18:27:44
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answer #2
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answered by Puggy 7
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sinx = -√2/2
sinx=-1/rt2
x=sin^-1(-1/rt2)
=sin^-1(-pi/4)
x=npi+(-1)^n(-pi/4) n element of Z
cos2theta = sinx at interval [0, 2pi]
cos2theta=sinx
=cos(pi/2-x)
2tehta=pi/2-x
2theta=2npi+/-x
theta=npi+/-x/2
cos2theta/3= -1 at interval [0, 2pi]
cos2theta/3=cos^-1pi
2theta/3=2npi+/-pi
theta=3npi+/-3pi/2
For this next problem, use a calculator to solve it, correct to our decimal places on the interval [ 0, 2pi]
3cossquaredx-8cosx-3=0
cosx=[8+/-rt(64+36)]/6
cosx=(8+10)/6=3 not acceptable as cosx max=1
cosx=8-10/3=-2/3
x=cos^-1(-2/3)
2006-12-14 18:20:27
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answer #3
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answered by raj 7
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sin x = -√2/2 ----> x=225 or 315 degrees
cos2theta = sinx ----> theta=22,5 or 112,5 degrees
-----> x= 45 or 225 degrees
cos(2theta/3) = -1 -----> theta= 270 degrees
3cos^2(x)-8cosx-3=0
(3cosx+1)(cosx-3)=0
3cosx+1=0 or cosx-3=0
cosx=-1/3 or cosx=3 (impossible)
so cosx=-1/3
x= 109,4712 or 250,5288 degrees
solved!!!
2006-12-14 18:17:13
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answer #4
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answered by fortman 3
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There are several important angles and values of sin and cos that are worth remembering.
One is :
sin theta = cos theta = √2/2
I'd recommend drawing a little sketch of the unit circle and playing with these.
cos [(2/3)*theta] = -1,
first, on the interval [0,2pi] there's only one angle where the cosine is -1,
2006-12-14 18:13:24
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answer #5
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answered by modulo_function 7
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Remember these identities. tanx = 1/cotx tanx = sinx/cosx csc^2x = 1 + cot^2x csc^2x = 1/sin^2x sin(2x) = 2sinxcosx Proving: = tanx + cotx = (1/cotx) + cotx = (1 + cot^2x) / cotx = csc^2x / cotx = csc^2x * tanx = 1/sin^2x * sinx/cosx = 1/sinxcosx * 2/2 = 2/(2sinxcosx) = 2/sin(2x) (Verified) Hope this helps!
2016-05-24 18:43:07
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answer #6
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answered by Anonymous
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