Question #d3a7a

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Question #d3a7a

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Answer 1 · 2016-11-29T11:01:03

Solution: $x = {31.7}^{0} (1 d p) and x = {121.7}^{0} (1 d p)$ $x=31.7^0(1dp) and x=121.7^0(1dp)$

Explanation:

$sin2x=2cos2x or (sin2x)/(cos2x)=2 or tan2x=2 or 2x= tan^-1 2 =63.4349^0:. x=31.7^0(1dp)$ and since $tan(pi+2x)=tan2x=2 :. 2x=180+63.4349=243.439^0:.x=121.7(1dp)$

Solution: $x=31.7^0(1dp) and x=121.7^0(1dp)$ [Ans]

Answer 2 · 2016-11-29T11:05:56

$sin2x=2cos2x$

$=>tan2x=2=tan^-1(2)=alpha$

So $2x=npi+alpha," where " n in ZZ$

$x=(npi)/2+alpha/2=(npi)/2+31.7^@$

by calculator, $alpha =tan^-1(2)= 63.4^@$

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