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प्रश्न
Solve the following equations:
sin 2θ – cos 2θ – sin θ + cos θ = θ
उत्तर
sin 2θ – cos 2θ – sin θ + cos θ = θ
`2cos ((2theta + theta)/2) sin ((2theta - theta)/2) - 2 sin ((2theta + theta)/2) sin ((theta - 2theta)/2)` = 0
`2cos ((3theta)/2) * sin (theta/2) - 2sin ((3theta)/2) sin (- theta/2)` = 0
`2cos ((3theta)/2) * sin (theta/2) + 2sin ((3theta)/2) sin (theta/2)` = 0
`2sin theta/2 [cos ((3theta)/2) + sin ((3theta)/2)]` = 0
`2 sin theta/2` = 0 or `cos ((3theta)/2) + sin ((3theta)/2)` = 0
`sin theta/2` = 0 or `cos ((3theta)/2) = - sin ((3theta)/2)`
`sin theta/2` = 0 or `(sin ((3theta)/2))/(cos ((3theta)/2))` = – 1
`sin theta/2` = 0 or `tan ((3theta)/2)` = – 1
To find the general solution of `sin theta/2` = 0
The general solution is
`theta/2` = nπ, n ∈ Z
θ = 2nπ, n ∈ Z
To find the general solution of `tan ((3theta)/2)` = – 1
`tan ((3theta)/2)` = – 1
`tan ((3theta)/2) = tan (pi - pi/4)`
`tan ((3theta)/2) = tan ((4pi - pi)/4)`
`tan ((3theta)/2) = tan ((3pi)/4)`
The general solution is
`(3theta)/2 = "n" + pi/4`, n ∈ Z
θ = `(2"n"pi)/3 + (2pi)/(3 xx 4)`, n ∈ Z
θ = `(2"n"pi)/3 + pi/6`, n ∈ Z
∴ The required solutions are
θ = 2nπ, n ∈ Z
or
θ = `(2"n"pi)/3 + pi/6`, n ∈ Z
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