# Calculate tgα if:2cos2α+4√3sinα=5 α(π /2; π )

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`we have`

`2cos(2alpha)+4sqrt(3)sin(alpha)=5` (i)

we know

`cos(2alpha)=1-2sin^2(alpha)`

Thus (i) reduces to

`2(1-2sin^2(alpha))+4sqrt(3)sin(alpha)=5` (ii)

Let us assume `sin(alpha)=t` ,so above equation reduces to

`2-4t^2+4sqrt(3)t-5=0`

`-4t^2+4sqrt(3)t-3=0`

`4t^2-4sqrt(3)t+2=0`

Also we know

`t=(4sqrt(3)+-sqrt(48-32))/8`

`t=(4sqrt(3)+-4)/8`

`t=(sqrt(3)+-1)/2`

`t=1.366`

`t=.366`

t=1.366 is not possible because `-1<=sin(alpha)<=1`

`sin(alpha)=.366`

`alpha=npi+(-1)^n(21.47)`

`n=1,then`

`alpha=158.53^o in(pi/2,pi)`

`` Ans.