For which values a, the equation \(\displaystyle{a}{\sin{{x}}}+{\left({a}+{1}\right)}{{\sin}^{{2}}{\frac{{{x}}}{{{2}}}}}+{\left({a}-{1}\right)}{{\cos}^{{2}}{\frac{{{x}}}{{{2}}}}}={1}\)

tibukooinm

tibukooinm

Answered question

2022-03-25

For which values a, the equation
asinx+(a+1)sin2x2+(a1)cos2x2=1

Answer & Explanation

Leonardo Mcpherson

Leonardo Mcpherson

Beginner2022-03-26Added 13 answers

asinx+(a+1)sin2x2+(a1)cos2x2=1
asinx+a(sin2x2+cos2x2)(cos2x2sin2x2)=1
asinx+acosx=1
asinxcosx=(1a)
a1+a2sinx11+a2cosx=(1a)1+a2
sinxcosαcossinα=sinβ
Where α=cos1(a1+a2) and β=sin1(1a1+a2)
sin(xα)=sinβ
xα=2nπ+βx=2nπ+α+β
or
xα=(2n+1)πβx=(2n+1)π+αβ

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