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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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Trig Help
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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1. Find the area of the region bounded by f(x)=x^2 +6x+9 and g(x)=5(x+3). Show the integral used, the limits of integration and how to evaluate the integral. 2. Find the area of the region bounded by x=y^2+6, x=0 , y=6, and y=7.
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Calc
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trig
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