Question

what is the minimum value of 3 sin6x+3cos6x​

Answer 1

Answer:

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Step-by-step explanation:

Answer 2

Answer:

y= sin^6(x) + cos^6(x).

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

y=(sin^2x+cos^2x)(sin^4x+cos^4x-sin^2x.cos^2x)

y=(1).[ (sin^2x+cos^2x)^2–3.sin^2x.cos^2x]

y= 1 - 3 sin^2x.cos^2x

y = 1 -3(sin x.cos x)^2

y= 1- 3(1/2.sin2x)^2

y = 1 - (3/4).(sin 2x)^2

For minimum value of y , sin2x should be maximum , maximum value of sin 2x =1 or

x=45°.

Minimum value of y= 1 -(3/4).(1)^2

= 1 - 3/4 = 1/4 , Answer.

Step-by-step explanation:

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