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Solve that following equations :`sec x cos 5x+1=0,` `0 lt x le pi/2` find the value of x |
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Answer» `secx*cos5x+1=0` `(cos5x)/cosx=-1` `cos5x=-cosx` `cos5x+cosx=0` `2cos((5x+x)/2)*cos((5x-x)/2)=0` `2ocs3x*cos2x=0` `cos3x=0` `3x=pi/2,3/2pi` `x=pi/6,pi/2` `cos2x=0` `2x=pi/2` `x=pi/4`. |
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