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For the following exercises, find all solutions exactly on the interval $0 \le \theta < 2\pi$. 4. $2\sin \theta = -\sqrt{2}$ 7. $2\cos \theta = -\sqrt{2}$ 10. $\cot x + 1 = 0$ 5. $2\sin \theta = \sqrt{3}$ 8. $\tan \theta = -1$ 11. $4\sin^2 x - 2 = 0$ 6. $2\cos \theta = 1$ 9. $\tan x = 1$ 12. $\csc^2 x - 4 = 0$

          For the following exercises, find all solutions exactly on the interval $0 \le \theta < 2\pi$.
4. $2\sin \theta = -\sqrt{2}$ 
7. $2\cos \theta = -\sqrt{2}$
10. $\cot x + 1 = 0$
5. $2\sin \theta = \sqrt{3}$
8. $\tan \theta = -1$
11. $4\sin^2 x - 2 = 0$
6. $2\cos \theta = 1$
9. $\tan x = 1$
12. $\csc^2 x - 4 = 0$
        
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For the following exercises, find all solutions exactly on the interval 0 ≤θ < 2π.
4. 2sinθ = -√(2) 
7. 2cosθ = -√(2)
10. cot x + 1 = 0
5. 2sinθ = √(3)
8. tanθ = -1
11. 4sin^2 x - 2 = 0
6. 2cosθ = 1
9. tan x = 1
12. csc^2 x - 4 = 0

Added by Josep R.

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Precalculus with Limits
Precalculus with Limits
Ron Larson 2nd Edition
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For the following exercises, find all solutions exactly on the interval 0<= heta <2pi . 10. cotx+1=0 For the following exercises,find all solutions exactly on the interval0<<2. 4.2sin0=-V2 7.2cos=-V2 10.cotx+1=0 5.2sin=3 8.tan=-1 6.2cos=1 9.tanx=1 11.4sin2x-2=0 12.cscx-4=0
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Transcript

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00:02 So if we model this equation as a quadratic, in other words, if we say we have x squared plus 2x minus 1 is equal to 0, minus 1 is equal to 0, and remember that we're setting x equal to tangent of omega.
00:21 Well, if we use the quadratic formula here, negative 2 plus or minus the square root of 4 minus 4 times negative 1 times 1, all over 2.
00:34 Gives us this gives value of negative 1 plus or minus plus or minus this gives us this gives root 8 so root 8 can be and be rewritten as as root 4 times root 2 and so root 4 it comes 2 so we have 2 root 2 divided by 2 this gives us plus or minus root 2 and so now we have that that tangent of omega is equal to negative 1 plus square root of 2 and tangent of omega is equal to negative 1, negative 1, and then we have minus root 2.
01:13 Well, we know that the square root of 2 is larger than 1.
01:20 So if we have negative 1 plus a root 2, this is going to be positive.
01:25 And then, so let's write this here.
01:26 So this will be positive.
01:27 And a negative 1 minus negative root 2 is going to be negative.
01:30 So when we're solving for our first value here, we're going to have that omega, omega is going to be equal to the tangent or tangent inverse...
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