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49. We have,

sin 2x + cosx = 0.

2 sin cosx + cosx = 0 ( $?$ sin 2x = 2 sin xcosx)

cosx (2 sin x + 1) = 0.

cosx = 0 or 2 sinx + 1 = 0.

x = (2n + 1) $\frac{\pi }{2}$, n∈z or sin x = $\frac{1}{2}$ = -sin $\frac{\pi }{6}$ = sin  π + $\frac{\pi }{6}$= sin $\frac{7\pi }{6}.$

x = (2n + 1) $\frac{\pi }{2}$, x∈z or x= nπ + (-1)ⁿ $\frac{7\pi }{6},$ n∈z.

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48. We have,

cos 3x + cosx-cos 2x = 0.

(cos 3x + cosx) cos 2x = 0

Using cos A + cos B = 2 cos $\frac{A+B}{2}$ cos $\frac{A-B}{2}$

2 cos $\left(\frac{3x+x}{2}\right)$ cos $\left(\frac{3x-x}{2}\right)$ - cos 2x = 0.

2 cos $\frac{4x}{2}$ cos $\frac{2x}{2}$ - cos 2x = 0

2 cos 2xcosx - cos 2x = 0

cos 2x. (2 cosx - 1) = 0.

cos 2x = 0 or 2 cosx -1 = 0.

2x = (2n + 1) $\frac{\pi }{2}$, n∈z or cosx = $\frac{1}{2}$ = cos $\frac{\pi }{3}$

x = (2n + 1) $\frac{\pi }{4}$, n∈z or x = 2nx± $\frac{\pi }{3}$, n∈z.

52. Kindly go through the solution

47. We have,

cos 4x = cos 2x.

⇒ cos 4x-cos 2x = 0.

⇒ -2 sin $\left(\frac{4x+2x}{2}\right)$ sin $\left(\frac{4x-2x}{2}\right)$ = 0.

⇒sin $\frac{6x}{2}$ sin $\frac{2x}{2}$ = 0

⇒sin 3x sin x = 0.

∴sin 3x = 0 or sin x = 0

3x = nπ or x = nπ, n∈z,

⇒x = $\frac{n\pi }{3}$ or x = nπ, n∈z

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51. Given, sin²x + cos²y = 1. 

Differentiating w r t ‘x’ we get,

$\frac{d}{dx}$ (sin²x + cos²y) $\frac{d}{dx}1.$

$\Rightarrow \frac{d}{dx}si{n}^{2}x+\frac{d}{dx}co{s}^{2}y=0$

$\Rightarrow 2sinx\frac{dsinx}{dx}+2cosy\frac{dcosy}{dx}=0$

= 2sin x cos x + 2 cos y   (- sin y) $\frac{dy}{dx}=0$

= sin 2x- sin 2y $\frac{dy}{dx}$ = 0

46. We have cosec x = 2 .i e, cosec x = (-)ve,

So, the principal solution lies in IIInd and IVth quadrent.

Now, cosec x = -2 = -cosec $\frac{\pi }{6}$ = cosec $(\pi +\frac{\pi }{6}$= cosec $\left(2\pi -\frac{\pi }{6}\right)$

So the principal solution are x = $\left(\pi +\frac{\pi }{6}\right)$ and $\left(2\pi -\frac{\pi }{6}\right)$

= $\left(\frac{6\pi +\pi }{6}\right)$ and $\left(\frac{12\pi -\pi }{6}\right)$

= $\frac{7\pi }{6}$ and $\frac{11\pi }{6}.$

As cosec x = cosec $\frac{7\pi }{6}$ ⇒sin x = sin $\frac{7\pi }{c}$ $\left[?cosecx=\frac{1}{sinx}\right]$

The general solution has the form,

x = nπ + (-1)n $\frac{7\pi }{6}$ , n∈z.

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Disclaimer: This info. is taken from the official website and may differ.

50. Given, sin²y + cos xy = p

Differentiating w r t ‘x’ we get,

$\frac{d}{dx}\left(si{n}^{2}y+cosxy\right)=\frac{d}{dx}$

= $\frac{d}{dx}si{n}^{2}y+\frac{d}{dx}cosxy=0$

$\Rightarrow 2siny\frac{d}{dx}(siny)+(-sinxy)\frac{d}{dx}(xy)=0.$

$\Rightarrow 2sinycosy\frac{dy}{dx}-sinxy\left[x\frac{dy}{dx}+y\right]=0$

$\Rightarrow 2sin2y\frac{dy}{dx}-xsinxy\frac{dy}{dx}-ysinxy=0$

{Qsin 2x = 2sin x cos x}

$\Rightarrow \frac{dy}{dx}[sin2y-xsinxy]=ysinxy.$

$\Rightarrow \frac{dxy}{dx}=\frac{ysinxy}{sin2y-xsinxy}.$

45. We have, cot x = -√3  i.e., cot x is negative

So, the principal solution lies in II^nd and IV^th quadrant.

So the principal solution are a = $\left(\pi -\frac{\pi }{6}\right)$ and $\left(2\pi -\frac{\pi }{6}\right)$

= $\left(\frac{6\pi -\pi }{6}\right)$ and $\left(\frac{12\pi -\pi }{6}\right)$

= $\frac{5\pi }{6}$ and $\frac{11\pi }{6}.$

As cot x = cot $\frac{5\pi }{6}$ tan x = tan $\frac{5\pi }{6}$ $\left[?cotx=\frac{1}{tanx}\right]$

The general solution has the form.

x = n+ $\frac{5\pi }{6}$ , nz

49. Given,  x³ + x²y + xy² + y³ = 81.

Differentiating w r t ‘x’ we get,

$\frac{d}{dx}\left(x^3+x^{2}y+x{y}^2+y^3\right)$ = $\frac{d(81)}{dx}$

$\Rightarrow \frac{d{x}^3}{dx}+\frac{d}{dx}{x}^{2}y+\frac{d}{dx}x{y}^2+\frac{d}{dx}{y}^3=0$

$\Rightarrow 3x^2+x^2\frac{dy}{dx}+y\frac{d{x}^2}{dx}+\frac{xd{y}^2}{dx}+y^2\frac{dx}{dx}+3y^2\frac{dy}{dx}=0$

$\Rightarrow 3x^2+x^2\frac{dy}{dx}+2xy+2xy\frac{dy}{dx}+y^2+3y^2\frac{dy}{dx}=0.$

$\Rightarrow \left(x^2+2xy+3y^2\right)\frac{dy}{dx}$= - (3x² + 2xy + y²)

$\frac{dy}{dx}=\frac{-\left(3x^2+2xy+y^2\right)}{\left(x^2+2xy+3y^2\right).}$

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48. Given,  x² + xy + y² = 100.

Differentiating w r t ‘x’ we get,

$\frac{d}{dx}\left(x^2+xy+y^2\right)=\frac{d}{dx}(100)$

$\Rightarrow 2x+x\frac{dy}{dx}+y\frac{dx}{dx}+2y\frac{dy}{dx}=0.$

$\Rightarrow x\frac{dy}{dx}+2y\frac{dy}{dx}=-2x-y$

$\Rightarrow \frac{dy}{dx}=\frac{-(2x+y)}{(x+2y)}$

43. tanx = √3

We have, tan x = √3

Since tan x is (+) ve the principal solution lies in I^st and III^nd quadrant

Now, tan x = √3 = tan $\frac{\pi }{3}$ . = tan $\left(\pi +\frac{\pi }{3}\right)$

Principal solution are x = $\frac{\pi }{3}$  and $\left(\pi +\frac{\pi }{3}\right)$

        $\frac{\pi }{3}$ =  and $\left(\frac{3\pi +\pi }{3}\right)$

        $\frac{\pi }{3}$ = and $\frac{4\pi }{3}$ .

As tan x = tan $\frac{\pi }{3}$

The general solution is.

x = np + $\frac{\pi }{3}$ , n∈z

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Disclaimer: This info. is taken from the official website and may differ.

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47. Given, xy + y² = tan x + y  Differentiating w r t x we get,

$\frac{d}{dx}\left(xy+y^2\right)=\frac{d}{dx}(tanx+y)$

$\Rightarrow x\frac{dy}{dx}+y\frac{dx}{dx}+\frac{d{y}^2}{dx}=d{x}^{2}x+\frac{dy}{dx}$

$\Rightarrow x\frac{dy}{dx}+2y\frac{dy}{dx}-\frac{dy}{dx}=se{n}^{2}x-y$

$\Rightarrow (x+2y-1)\frac{dy}{dx}=se{c}^{2}x-y$

$\Rightarrow \frac{dy}{dx}=\frac{si{n}^{2}x-y}{x+2y-1.}$

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Source: Official site and may vary.