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When x=0:tan(0)+sec(2⋅0)=1 so x=0 is a solution, because tan(0)=0andcos(0)andsec(0)=1. There is another solution at x=67.5 degrees. See below: If y=sin(x)ortan(x).cos(2x)=1−2sin2(x)=1−2y2,sec(2x)=11−2sin2(x),cos(x)=1−y2.sin=opphyp=y1soadj=1−y2.tan(x)=oppadj=y1−y2. Therefore, tan(x)+sec(2x)=y1−y2+11−2y2=1. y1−y2=1−11−2y2=1−2y2−11−2y2=−2y21−2y2. y=0 is a solution (sin(x)=0sox=0)and: 11−y2=−2y1−2y2, 1−y2=2y2−12y. Squaring: 1−y2=4y4−4y2+14y2,4y2−4y4=4y4−4y2+1,8y4−8y2+1=0. So y2=(8±64−3216=12±24 Not exactly what you’re looking for? Ask My Question This is helpful 84
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1secθ−tanθ+1secθ+tanθ
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