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posted by Mike
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trig
find the exact solutions 2cos^2x+3sinx=0 the way it stands, that is a "nasty" question. Are you sure the second term isn't 2sin(2x) ? no, its as i wrote it. then it's got me stymied, I must be missing something rather obvious,
asked by Devon on April 16, 2007 
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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.
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I was given 21 questions for homework and I can't get the last few no matter how hard and how many times I try. 17. Sinx1/sinx+1 = cos^2x/(sinx+1)^2 18. Sin^4x + 2sin^2xcos^2x + cos^4x = 1 19. 4/cos^2x  5 = 4tan^2x  1 20. Cosx
asked by Kim on December 4, 2012 
help, bobpursely
Please take a look at DEVON's trig equation at 9:14 I must be missing something obvious. If the second had been 3sin 2x it would be a straight forward question. The way it stands I was able to "reduce" it to the equation 8sin^4 x
asked by Reiny on April 16, 2007 
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
asked by Maggie on February 2, 2015 
math
y= 2sin^2 x y=1 sinx find values of x inthe interval 0
asked by Candy on December 14, 2006 
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Find the dervatives: 1. f(x)=(3x+1)e^x^2 2. y = e^(sin x) ln(x) 3. f(x)=(x^(2)+x)^23 4. f(x)=sin(2x)/cosx 5. square root of x/(3x+1) 6. f(x)=sin^4(3x=1)sin(3x+1) 7. x+y=cos(xy) The answers I got: 1.
asked by Anonymous on October 15, 2012 
math
the problem is 2cos^2x + sinx1=0 the directions are to "use an identity to solve each equation on the interval [0,2pi). This is what i've done so far: 2cos^2x+sinx1=0 2cos^2x1+sinx=0 cos2x + sinx =0 1  2sin^2x + sinx = 0
asked by alex on April 7, 2008 
PreCalculus
Solve the equation on the interval [0,2pi). 2sin^2x3sinx+1=0 (2sinx+1)(sinx+1) I don't think I did the factoring correctly. When I multiply it out to double check I get 2sin^2x+3sinx+1
asked by Jan on October 27, 2017