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The number of solution(s) of the trigonometric equation sec^2x+cosec^2x-sin2x-3=0, in [0, 4pi], is |
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Answer» 2 `1+tan^2x+1 + cot^2x=1+sin2x+2` `2+tan^2x+cot^2x =(sin x+cosx)^2 +2` Range of LHS is `[4,oo]` Range of RHS is [2,4] So SOLUTIONWILL exist when sin2x=1 and tanx=cotx= `pm`1 `2x=x/2(4n+1)` `x=pi/4,(5pi)/4,(9pi)/4,(13pi)/4` at these x the VALUE of tan x = cot x =1 so number of solution = 4 |
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