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# help trig

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If sin(2x) = 2/3, then find sin(x)^6 + cos(x)^6.

Apr 13, 2022

#2
+2399
+1

sin(2x) = 2/3

2sin(x)cos(x) = 2/3

sin(x)cos(x) = 1/3

sin^2(x)cos^2(x) = 1/9

3sin^2(x)cos^2(x) = 1/3

sin^2(x) + cos^2(x) = 1

sin^6(x) + 3sin^4(x)cos^2(x) + 3sin^2(x)cos^4(x) + cos^6(x) = 1

sin^6(x) + cos^6(x) + 3sin^2(x)cos^2(x)(sin^2(x) + cos^2(x)) = 1

sin^6(x) + cos^6(x) + 3sin^2(x)cos^2(x)(1) = 1

sin^6(x) + cos^6(x) + 3sin^2(x)cos^2(x) = 1

sin^6(x) + cos^6(x) + 1/3 = 1

sin^6(x) + cos^6(x) = 2/3

=^._.^=

Apr 13, 2022

#1
+115
-2

I believe it 9/6

Apr 13, 2022
#2
+2399
+1

sin(2x) = 2/3

2sin(x)cos(x) = 2/3

sin(x)cos(x) = 1/3

sin^2(x)cos^2(x) = 1/9

3sin^2(x)cos^2(x) = 1/3

sin^2(x) + cos^2(x) = 1

sin^6(x) + 3sin^4(x)cos^2(x) + 3sin^2(x)cos^4(x) + cos^6(x) = 1

sin^6(x) + cos^6(x) + 3sin^2(x)cos^2(x)(sin^2(x) + cos^2(x)) = 1

sin^6(x) + cos^6(x) + 3sin^2(x)cos^2(x)(1) = 1

sin^6(x) + cos^6(x) + 3sin^2(x)cos^2(x) = 1

sin^6(x) + cos^6(x) + 1/3 = 1

sin^6(x) + cos^6(x) = 2/3

=^._.^=

catmg Apr 13, 2022
#3
+115
-2

so my answe not correct hm?

Kakashi  Apr 13, 2022
#4
+2399
0

I don't this so.

=^._.^=

catmg  Apr 13, 2022
#5
+124701
0

Excellent, catmg   !!!!

CPhill  Apr 14, 2022