Question

Solve for x :
2^sin^2x+2^cos^2x=2√2
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Answer 1

Using AM-GM inequality ( since 2n>02n>0 for all realnn)

2sin2x+2cos2x2≥2sin2x+cos2x−−−−−−−−√2sin2x+2cos2x2≥2sin2x+cos2x

Now sin2x+cos2x=1,2sin2x+cos2x−−−−−−−−√=2–√sin2x+cos2x=1,2sin2x+cos2x=2

Therefore 2sin2x+2cos2x2≥2–√2sin2x+2cos2x2≥2

2sin2x+2cos2x≥22

Answer 2

Answer:

At x=45 degree the above condition will satisfy

Step-by-step explanation: