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3^sin²x+3^sin²x=4 then |sinx|+|cosx|

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ANSWER

\frac{ \sqrt{2}  + <klux>1</klux>}{ \sqrt{<klux>3</klux>} }

Step-by-step explanation:

6 {sin}^{2}x \:  = 4

so \: sinx =  \sqrt{ \frac{2}{3} }

cosx =  \sqrt{ \frac{1}{3} }

therefore |sinx| + |cosx| = 2*1/2 + 1/ 3*1/2



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