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The general solution of sin x − 3 sin2x + sin3x = cos x − 3 cos2x + cos3x is _________. Solution: sin x … If sec 4θ − sec 2θ = 2, then the general value of θ is __________. Solution: sec 4θ − sec 2θ = 2 ⇒ cos … If tan (cot x) = cot (tan x), then sin 2x = ___________. Solution: tan (cotx) = cot (tanx) ⇒ tan (cotx) = … If the solution for θ of cospθ + cosqθ = 0, p > 0, q > 0 are in A.P., then numerically the smallest …

Answers

Answered by friendlygirl4
1

Step-by-step explanation:

sec 4θ − sec 2θ = 2 ⇒ cos … If tan (cot x) = cot (tan x), then sin 2x = ___________. Solution: tan (cotx) = cot (tanx) ⇒ tan (cotx) = … If the solution for θ of cospθ + cosqθ = 0, p > 0, q > 0 are in A.P., then numerically the smallest …

Answered by Anonymous
32

Step-by-step explanation:

We have, (sinx+sin3x)−3sin2x=(cosx+cos3x)−3cos2x

We have, (sinx+sin3x)−3sin2x=(cosx+cos3x)−3cos2x⇒2sin2xcosx−3sin2x=2cos2xcosx−3cos2x

We have, (sinx+sin3x)−3sin2x=(cosx+cos3x)−3cos2x⇒2sin2xcosx−3sin2x=2cos2xcosx−3cos2x⇒sin2x(2cosx−3)=cos2x(2cosx−3)

We have, (sinx+sin3x)−3sin2x=(cosx+cos3x)−3cos2x⇒2sin2xcosx−3sin2x=2cos2xcosx−3cos2x⇒sin2x(2cosx−3)=cos2x(2cosx−3)⇒sin2x=cos2x(Ascosx

We have, (sinx+sin3x)−3sin2x=(cosx+cos3x)−3cos2x⇒2sin2xcosx−3sin2x=2cos2xcosx−3cos2x⇒sin2x(2cosx−3)=cos2x(2cosx−3)⇒sin2x=cos2x(Ascosx

We have, (sinx+sin3x)−3sin2x=(cosx+cos3x)−3cos2x⇒2sin2xcosx−3sin2x=2cos2xcosx−3cos2x⇒sin2x(2cosx−3)=cos2x(2cosx−3)⇒sin2x=cos2x(Ascosx=3/2)

We have, (sinx+sin3x)−3sin2x=(cosx+cos3x)−3cos2x⇒2sin2xcosx−3sin2x=2cos2xcosx−3cos2x⇒sin2x(2cosx−3)=cos2x(2cosx−3)⇒sin2x=cos2x(Ascosx=3/2)⇒tan2x=1⇒tan2x=tanπ/4

We have, (sinx+sin3x)−3sin2x=(cosx+cos3x)−3cos2x⇒2sin2xcosx−3sin2x=2cos2xcosx−3cos2x⇒sin2x(2cosx−3)=cos2x(2cosx−3)⇒sin2x=cos2x(Ascosx=3/2)⇒tan2x=1⇒tan2x=tanπ/4⇒2x=nπ+

We have, (sinx+sin3x)−3sin2x=(cosx+cos3x)−3cos2x⇒2sin2xcosx−3sin2x=2cos2xcosx−3cos2x⇒sin2x(2cosx−3)=cos2x(2cosx−3)⇒sin2x=cos2x(Ascosx=3/2)⇒tan2x=1⇒tan2x=tanπ/4⇒2x=nπ+ 4

We have, (sinx+sin3x)−3sin2x=(cosx+cos3x)−3cos2x⇒2sin2xcosx−3sin2x=2cos2xcosx−3cos2x⇒sin2x(2cosx−3)=cos2x(2cosx−3)⇒sin2x=cos2x(Ascosx=3/2)⇒tan2x=1⇒tan2x=tanπ/4⇒2x=nπ+ 4π

We have, (sinx+sin3x)−3sin2x=(cosx+cos3x)−3cos2x⇒2sin2xcosx−3sin2x=2cos2xcosx−3cos2x⇒sin2x(2cosx−3)=cos2x(2cosx−3)⇒sin2x=cos2x(Ascosx=3/2)⇒tan2x=1⇒tan2x=tanπ/4⇒2x=nπ+ 4π

We have, (sinx+sin3x)−3sin2x=(cosx+cos3x)−3cos2x⇒2sin2xcosx−3sin2x=2cos2xcosx−3cos2x⇒sin2x(2cosx−3)=cos2x(2cosx−3)⇒sin2x=cos2x(Ascosx=3/2)⇒tan2x=1⇒tan2x=tanπ/4⇒2x=nπ+ 4π ,n∈Z

We have, (sinx+sin3x)−3sin2x=(cosx+cos3x)−3cos2x⇒2sin2xcosx−3sin2x=2cos2xcosx−3cos2x⇒sin2x(2cosx−3)=cos2x(2cosx−3)⇒sin2x=cos2x(Ascosx=3/2)⇒tan2x=1⇒tan2x=tanπ/4⇒2x=nπ+ 4π ,n∈Zx=

We have, (sinx+sin3x)−3sin2x=(cosx+cos3x)−3cos2x⇒2sin2xcosx−3sin2x=2cos2xcosx−3cos2x⇒sin2x(2cosx−3)=cos2x(2cosx−3)⇒sin2x=cos2x(Ascosx=3/2)⇒tan2x=1⇒tan2x=tanπ/4⇒2x=nπ+ 4π ,n∈Zx= 2

We have, (sinx+sin3x)−3sin2x=(cosx+cos3x)−3cos2x⇒2sin2xcosx−3sin2x=2cos2xcosx−3cos2x⇒sin2x(2cosx−3)=cos2x(2cosx−3)⇒sin2x=cos2x(Ascosx=3/2)⇒tan2x=1⇒tan2x=tanπ/4⇒2x=nπ+ 4π ,n∈Zx= 2nπ

We have, (sinx+sin3x)−3sin2x=(cosx+cos3x)−3cos2x⇒2sin2xcosx−3sin2x=2cos2xcosx−3cos2x⇒sin2x(2cosx−3)=cos2x(2cosx−3)⇒sin2x=cos2x(Ascosx=3/2)⇒tan2x=1⇒tan2x=tanπ/4⇒2x=nπ+ 4π ,n∈Zx= 2nπ

We have, (sinx+sin3x)−3sin2x=(cosx+cos3x)−3cos2x⇒2sin2xcosx−3sin2x=2cos2xcosx−3cos2x⇒sin2x(2cosx−3)=cos2x(2cosx−3)⇒sin2x=cos2x(Ascosx=3/2)⇒tan2x=1⇒tan2x=tanπ/4⇒2x=nπ+ 4π ,n∈Zx= 2nπ +

We have, (sinx+sin3x)−3sin2x=(cosx+cos3x)−3cos2x⇒2sin2xcosx−3sin2x=2cos2xcosx−3cos2x⇒sin2x(2cosx−3)=cos2x(2cosx−3)⇒sin2x=cos2x(Ascosx=3/2)⇒tan2x=1⇒tan2x=tanπ/4⇒2x=nπ+ 4π ,n∈Zx= 2nπ + 8

We have, (sinx+sin3x)−3sin2x=(cosx+cos3x)−3cos2x⇒2sin2xcosx−3sin2x=2cos2xcosx−3cos2x⇒sin2x(2cosx−3)=cos2x(2cosx−3)⇒sin2x=cos2x(Ascosx=3/2)⇒tan2x=1⇒tan2x=tanπ/4⇒2x=nπ+ 4π ,n∈Zx= 2nπ + 8π

We have, (sinx+sin3x)−3sin2x=(cosx+cos3x)−3cos2x⇒2sin2xcosx−3sin2x=2cos2xcosx−3cos2x⇒sin2x(2cosx−3)=cos2x(2cosx−3)⇒sin2x=cos2x(Ascosx=3/2)⇒tan2x=1⇒tan2x=tanπ/4⇒2x=nπ+ 4π ,n∈Zx= 2nπ + 8π

We have, (sinx+sin3x)−3sin2x=(cosx+cos3x)−3cos2x⇒2sin2xcosx−3sin2x=2cos2xcosx−3cos2x⇒sin2x(2cosx−3)=cos2x(2cosx−3)⇒sin2x=cos2x(Ascosx=3/2)⇒tan2x=1⇒tan2x=tanπ/4⇒2x=nπ+ 4π ,n∈Zx= 2nπ + 8π ,n∈Z

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