given x=2cos(t)+tsin(t) and y=2sin(t)-tcost(t), with

the problem is2cos^2x + sinx-1=0 the directions are to "use an identity to solve each equation on the interval [0,2pi). This is
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Which expression is equivalent to sin(3x) + sin x?A. 2cos(2x)sin x B. 2sin(2x)sin x C. -2sin(2x)cos x D. 2sin(2x)cos x E.
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Which expression is equivalent to sin(3x) + sin x?A) 2cos(2x)sin x B) 2sin(2x)sin x C) -2sin(2x)cos x D) 2sin(2x)cos x E)
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Which expression is equivalent to sin(3x) + sin x?A. 2cos(2x)sin x B. 2sin(2x)sin x C. -2sin(2x)cos x D. 2sin(2x)cos x E.
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I am given a vector function <2cos(t) + cos(2t), 2sin(t) + sin(2t), 0>(in other words) : (2cos(t) + cos(2t))i + (2sin(t) +
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What am I doing wrong?Equation: sin2x = 2cos2x Answers: 90 and 270 .... My Work: 2sin(x)cos(x) = 2cos(2x) sin(x) cos(x) =
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given x=2cos(t)+t*sin(t) and y=2sin(t)-t*cost(t), with t:[pi, 2pi]find all critical values. (4 answers)
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Verify the identities.Cos^2x - sin^2x = 2cos^2x - 1 When verifying identities, can I work on both side? Ex. 1 - sin^2x - sin^2x
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how does thederivative of -2xcos(x^2)= -4xcos(x^2) -2sin(x^2) and not -4x^2cos(x^2) -2sin(x^2)?
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given x=2cos(t)+t*sin(t) and y=2sin(t)-t*cost(t), with

given x=2cos(t)+tsin(t) and y=2sin(t)-tcost(t), with