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Question Number 45280 by peter frank last updated on 11/Oct/18

solve the simultaneous equation

a) \sin (x + y) = \frac{1}{\sqrt{2}}

\cos 2 x = \frac{- 1}{2} for x and y ranging from 0 to 360 inclusive

b) if siny + cosx = x show that \frac{d^{2} y}{{dx}^{2}} = \frac{x}{{(2 - x^{2})}^{\frac{3}{2}}}

Answered by tanmay.chaudhury50@gmail.com last updated on 11/Oct/18

s i n (x + y) = \frac{1}{\sqrt{2}} = s i n (π - \frac{π}{4})

x + y = \frac{3 π}{4}

c o s 2 x = - \frac{1}{2} = c o s (π - \frac{π}{3})

2 x = \frac{2 π}{3} x = \frac{π}{3} y = \frac{3 π}{4} - \frac{π}{3} = \frac{5 π}{12}

x = \frac{π}{3} y = \frac{5 π}{12}

Commented by peter frank last updated on 11/Oct/18

thank you sir . pls help qn 2

Answered by tanmay.chaudhury50@gmail.com last updated on 12/Oct/18

\frac{d}{d x} (\frac{d y}{d x}) = \frac{x}{{(2 - x^{2})}^{\frac{3}{2}}}

\int d (\frac{d y}{d x}) = \int \frac{x d x}{{(2 - x^{2})}^{\frac{3}{2}}}

t^{2} = 2 - x^{2} 2 t d t = - 2 x d x

R H S = \int \frac{- t d t}{t^{3}}

= \int - t^{- 2} d t

= - 1 \times \frac{t^{- 1}}{- 1} = \frac{1}{t} = \frac{1}{\sqrt{2 - x^{2}}}

\frac{d y}{d x} = \frac{1}{\sqrt{2 - x^{2}}}

\int d y = \int \frac{d x}{\sqrt{2 - x^{2}}} x = \sqrt{2} s i n θ d x = \sqrt{2} c o s θ d θ

y = \int \frac{\sqrt{2} c o s θ}{\sqrt{2} c o s θ} d θ

y = θ + c

y = {s i n}^{- 1} (\frac{x}{\sqrt{2}}) + c

s i n (y - c) = \frac{x}{\sqrt{2}}

x = \sqrt{2} s i n (y - c)

S O p l s c h e c k t h e q u e s t i o n \dots .

Commented by peter frank last updated on 12/Oct/18

its seem there are something wrong . thanks sir