Math, asked by rajeshsinghbharadwaj, 9 months ago

y =
Cos2x + 3 Sinx find range​

Answers

Answered by senboni123456
13

Step-by-step explanation:

Given,

y =  \cos(2x)  + 3 \sin(x)

 =  > y = 1 - 2 \sin^{2} (x)  + 3 \sin(x)

 =  > y =  - (2 \sin^{2} (x) - 3 \sin(x) - 1)

 =  > y =  - 2( \sin^{2} (x)   -  \frac{3}{2} \sin(x)   -  \frac{1}{2} )

 =  > y =  - 2(( \sin(x) -  \frac{3}{4})^{2}  - ( \frac{9}{16} +  \frac{1}{2}  ))

 =  > y = - 2( \sin(x) -  \frac{3}{4} ) ^{2} +  \frac{17}{8}

Now, we have,

 =  > ( \sin(x)  -  \frac{3}{4})^{2}  \geqslant 0

 =  >  - 2( \sin(x)  -  \frac{3}{4})^{2} \leqslant 0

 =  >   - 2( \sin(x)  -  \frac{3}{4})^{2}  +  \frac{17}{8} \leqslant  \frac{17}{8}

 =  > y  \leqslant  \frac{17}{8}

Hence the range is

(-infinity, 17/8]

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