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A: sin θ = 15cos2 θ = 1 - sin2θcos2θ = 1 -(15)2 cos2θ = 1 -125cos2θ =2425cos θ = 265
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- How many solutions on the interval [0, 2020π] does the equation 1/2 sin2x + 1 = sinx + cosx have?On Section 3.6, question 25. I do not understand what rule is used to convert derivative sin 3t to (cos 3t*3) or why the derivative for cos 5t is (-sin 5t *5)? So I can't figure out the problem. Can you help?Solve the equation 2 sin2 x + cos x = 1 on the interval [0, 2π).Use exact values where possible or give approximate solutions correct to four decimal places.