solve sin3x+sin2x-sinx=0
Answers
Answered by
5
sin3x+sin2x-sinx=0
sin4x=0
We know, sin0°=0
Comparing,
Sin4x=sin0°
4x=0
Therefore,
x=0
ThankYou
sin4x=0
We know, sin0°=0
Comparing,
Sin4x=sin0°
4x=0
Therefore,
x=0
ThankYou
shubham991:
galat kiya h
sin c + sin d....wale formula se hoga
no i think we should use sinc-sind
ek hi baat h
the formula is different for both sinc+sind and sinc-sind
are mtlb kch b use krlo...ans aa jyga
i did it but i didnt get answer so only i asked
I'm sorry if the answer is wrong
I wasn't sure about it
Answered by
18
here's a method that doesn't involve sum to product formula
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)
sinx + sin2x - sin3x = 0
sinx + 2sinx cosx + 4sin³x - 3sinx = 0
2sinx cosx + 4sin³x - 2sinx = 0
sinx cosx + 2sin³x - sinx = 0
sinx (cosx + 2sin²x - 1) = 0
sinx (cosx + 2(1 - cos²x) - 1) = 0
sinx (cosx + 2 - 2cos²x - 1) = 0
sinx (-2cos²x + cosx + 1) = 0
sinx (2cos²x - cosx - 1) = 0
sinx (cosx - 1)(2cosx + 1) = 0
sinx = 0 ==> x = kπ
cosx - 1 = 0 ==> x = kπ
2cosx + 1 = 0 ==> x = 2π/3 + 2kπ, 5π/3 + 2kπ
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)
sinx + sin2x - sin3x = 0
sinx + 2sinx cosx + 4sin³x - 3sinx = 0
2sinx cosx + 4sin³x - 2sinx = 0
sinx cosx + 2sin³x - sinx = 0
sinx (cosx + 2sin²x - 1) = 0
sinx (cosx + 2(1 - cos²x) - 1) = 0
sinx (cosx + 2 - 2cos²x - 1) = 0
sinx (-2cos²x + cosx + 1) = 0
sinx (2cos²x - cosx - 1) = 0
sinx (cosx - 1)(2cosx + 1) = 0
sinx = 0 ==> x = kπ
cosx - 1 = 0 ==> x = kπ
2cosx + 1 = 0 ==> x = 2π/3 + 2kπ, 5π/3 + 2kπ
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