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sin3θ = cos2θ find the most general values of θ satisfying the equatios? sinax + cosbx = 0 solve ?

sin3θ = cos2θ find the most general values of θ satisfying the equatios?


sinax + cosbx = 0 solve ?

Grade:11

1 Answers

Aman Bansal
592 Points
10 years ago

Dear Mayank,

sin (3x)
= sin(2x + x)
= sin(2x)cos(x) + cos(2x)sin(x)
= 2sin(x)cos(x)cos(x) + (1 - 2sin² (x) ) sin(x)
= 2sin(x)cos² (x) + sin(x) - 2sin³ (x)
= 2sin(x) ( 1 -sin² (x) ) + sin(x) - 2sin³ (x)
= 2sin(x) -2sin³ (x) + sin(x) - 2sin³ (x)
= -4sin³ (x) + 3sin(x)
= 3sin(x) - 4sin³ (x)

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