Question

the maximum value of 2sinA+3cosA​

Answer 1

Answer:

18/5

Step-by-step explanation:

first derivative of the function..,

f′(x)=4cosA−3sinA  

f′(x) will be zero when 4cosA = 3sinA

i.e tan A=4/3  => sin A=3/5 and cos A=4/5.

we hv 2(3/5)+3(4/5)

=>18/5

Answer 2

The maximum value of given expression is √13.

Step-by-step explanation:

Since we have given that

$2\sin A+3\cos A$

Maximum value of asin A+b cos A is given by

$\sqrt{2^2+3^2}\\\\=\sqrt{4+9}\\\\=\sqrt{13}$

So, the maximum value of given expression is √13.

# learn more:

3cotA=2 find 2sinA-3cosA/2sinA+3cosA

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