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Math Help
Hello! Can someone please check and see if I did this right? Thanks! :) Directions: Find the exact solutions of the equation in the interval [0,2pi] cos2x+sinx=0 My answer: cos2x+sinx=cos^2xsin^2x+sinx =1sin^2xsin^2x+sinx
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I need help solving for all solutions for this problem: cos 2x+ sin x= 0 I substituted cos 2x for cos^2xsin^2x So it became cos^2(x)sin^2(x) +sinx=0 Then i did 1sin^2(x)sin^2(x)+sinx=0 = 12sin^2(x)+sinx=0 = sinx(2sinx+1)=1
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I need to find all solutions of the given equations for the indicated interval. Round solutions to three decimal places if necessary. 1.) 3sin(x)+1=0, x within [0,2pi) 2.) 2sin(sq'd)(x)+cos(x)1=0, x within R 3.)
asked by Emily on June 21, 2009 
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Mathematics  Trigonometric Identities  Reiny, Friday, November 9, 2007 at 10:30pm (sinx  1 cos^2x) (sinx + 1  cos^2x) should have been (sinx  1 + cos^2x) (sinx + 1  cos^2x) and then the next line should be sin^2x + sinx 
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4. Find the exact value for sin(x+y) if sinx=4/5 and cos y = 15/17. Angles x and y are in the fourth quadrant. 5. Find the exact value for cos 165degrees using the halfangle identity. 1. Solve: 2 cos^2x  3 cosx + 1 = 0 for 0
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Several questions from my homework, any are appreciated. Thanks! Solve: 1) sinxtanx=0 2) 3csc^2x=4 Find all solutions within (0,2pi) 3) cos4x(cosx1)=0 4) cos4x7cos2x=8 Use inverse functions where necessary to find solutions
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Prove the following: [1+sinx]/[1+cscx]=tanx/secx =[1+sinx]/[1+1/sinx] =[1+sinx]/[(sinx+1)/sinx] =[1+sinx]*[sinx/(sinx+1)] =[sinx+sin^2x]/[sinx+1] =[sinx+(1cos^2x)]/[sinx+1] =??? This is where I'm stuck. Can someone help me.
asked by Anonymous on March 11, 2012 
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Can I please get some help on these questions: 1. How many solutions does the equation,2sin^2 2 θ = sin2θ have on the interval [0, 2pi]? 4? ...what about 4cos2θ = 8cos^2 2θ? 2. True or False: sin^2 4x = 1 has 8 solutions on
asked by mysterychicken on June 2, 2013