InterviewSolution
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InterviewSolution
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Solve the equation:`cos^2[pi/4(sinx+sqrt(2)cos^2x)]-tan^2[x+pi/4tan^2x]=1` |
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Answer» `cos^(2) [pi/4(sin x+sqrt(2) cos^(2) x)]-tan^(2) [x+pi/4 tan^(2) x]=1` `rArr sin^(2) [pi/4 (sin x+ sqrt(2) cos^(2) x)]+tan^(2) [x+pi/4 tan^(2) x]=0` It is possible only when `sin^(2) [pi/4 (sin x+sqrt(2) cos^(2) x)]=0`...(i) and `tan^(2) [x+pi/4 tan^(2) x]=0` ...(ii) `:. pi/4 (sin x+ sqrt(2) cos^(2) x) = n pi, n in I` or `sin x +sqrt(2) cos^(2) x=4n` This equation has solution only for `n=0`. Thus, `sin x+sqrt(2) cos^(2) x=0` i.e., `sqrt(2) sin^(2) x- sin x-sqrt(2)=0` or `(sin x-sqrt(2)) (sqrt(2) sin x +1) =0` `:. sin x= - 1/sqrt(2)` `rArr x=2 kpi-pi//4, k in Z` Also these values of x satisfy Eq. (ii), therefore , the general solution of given equation is given by `x=2kpi - pi/4, k in Z` |
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