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If `sintheta,1,cos2theta` are in G.P., then find the general values of `theta` |
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Answer» Correct Answer - `theta=npi+(-1)^(n-1) pi/2` `1^(2)= sin theta cos 2 theta` or `1- sin theta (1-2 sin^(2) theta)=0` or `2 sin^(3) theta- sin theta+1=0` or `(sin theta+1) (2 sin^(2) theta-2 sin theta+1)=0` or `sin theta=-1` the other factor gives imaginary roots `rArr theta=n pi +(-1)^(n) (-pi/2)` `=n pi-(-1)^(n) pi/2= n pi+(-1)^(n-1) pi/2` |
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