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Find the general solution of sin2θ=cos3θ .

Answer
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Hint: Convert the whole equation in terms of sin function only, using relation sin(π2x)=cosx with the term cos3θ . use the identity of sinCsinD to get the general solution, which is given as –

sinCsinD=2sinCD2cosC+D2

And use the following results as well

General solution of sinx=0 is x=nπ and cosx=0 is x=(2n+1)π2 , nz .

Complete step-by-step answer:
Here, we have –

sin2θ=cos3θ ……………………………………… (i)

As the equation (i) is consisting of two different functions sin and cos both. So, we can use following trigonometric identity to get the whole equation in terms of sin only as –

sin(π2x)=cosx …………………………. (ii)

Now, we can replace cos3θ by sin(π23θ) of the equation (i), using the equation (ii). So, we get the equation (i) as –

sin2θ=sin(π23θ) .

Transferring the term sin(π23θ) to the other side of the equation, we get –

sin2θsin(π23θ)=0 (iii)

Now, as we know the trigonometric identity of sinCsinD is given as –

sinCsinD=2sinCD2cosC+D2 (iv)

Now, on comparing the equation (iii) and (iv), we get –

C=2θ and D=(π23θ).

So, we can rewrite the equation (iii) with the help of equation (iv) as –

2sin(2θ(π23θ)2)cos(2θ+(π23θ)2)=02sin(2θπ2+3θ2)cos(2θ+π23θ2)=0

Or sin(5θπ22)cos(θ+π22)=0

sin(5θ2π4)cos(θ2+π4)=0………………………………(iv)

N0w, we know that two terms in a product will be zero, if one of them will be 0. So, we can get two expressions from the equation (iv) as –

sin(5θ2π4)=0 or cos(θ2+π4)=0.

Case 1: sin(5θ2π4)=0

So, the general solution of the equation sin(5θ2π4)=0 can be given with the help of relation (v) as –

5θ2π4=nπ .

Adding π4 to both sides of the equation, we get –

5θ2π4+π4=nπ+π45θ2=nπ+π4

Multiplying the whole equation by ‘25’ , we get –

5θ2×25=(nπ+π4)×25θ=2nπ5+2π20


Or θ=2nπ5+π10…………………………… (vi)


Where, nz .


Case 2: cos(θ2+π4)=0


Hence, the general solution of the equation cos(θ2+π4)=0 by using the equation (vii). We get –


θ2+π4=(2n+1)π2 .


Subtracting π4 from the both sides of the equation, we get –


θ2+π4π4=(2n+1)π2π4θ2=(2n+1)π2π4


Multiplying the whole equation by ‘-2’, we can rewrite the above equation as –


θ2(2)=((2n+1)π2π4)×(2)θ=(2n×π2+π2π4)(2)θ=(nπ+π4)(2)

θ=2nππ2…………………………………….. (viii)


Where nz

Hence, we can get the general solution of the equation sin2θ=cos3θ , by

using the equations (vi) and (viii) as –

θ=2nπ5+π10,2nππ2


Where, nz.


Note: Another approach to solve the question would be given as –

sin2θ=cos3θcos(π22θ)=cos3θ

Use a general solution for the relation cosx=cosy i.e. x=2nπ±y . So, we get –

π22θ=2nπ±3θ

Case 1:

π22θ=2nπ±3θ5θ=π22nπθ=π102nπ5,nz.

Case 2:

π22θ=2nπ3θθ=2nππ2

Where, nz

So, it can be another approach. One may write the relation sin2θ=cos3θ as sin2θ=sin(π23θ). And hence, use the general solution for sinx=siny , which is given as –

x=nπ+(1)ny .

Where, nz.

So, one may observe that the representation of the solution of the same equation is different but the solutions on putting values of ‘n’ will be the same. Hence, do not confuse, if your answer is not matching with the given solution, just put the values of n and hence, try to verify your relation.