Class 11th

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.

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.

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

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.

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

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

41. L.H.S. = cos 4x.

= cos 2 (2x)

= 1 – 2 Sin² (2x) [ cos 2x = 1 – 2 Sin²x]

= 1 – 2 [2 sin xcosx]² [ sin 2x = 2 sin xcos x]

= 1 – 2 [4 sin²xcos²x]

= 1 – 8 sin²xcos²x

= R.H.S.

40. $$L.H.S. = tan 4x

We know that,

$tan2=\frac{2tan}{1-ta{n}^2}.$ , we can write

$L.H.S=tan2(2x)=\frac{2tan2x}{1-ta{n}^{2}2x.}$

$=\frac{2\left(\frac{2tanx}{1-ta{n}^{2}x}\right)}{1-{\left(\frac{2tanx}{1-ta{n}^{2}x}\right)}^2}$

$=\frac{\frac{4tanx}{\left(1-ta{n}^{2}x\right)}}{\frac{{\left(1-ta{n}^{2}x\right)}^2-4ta{n}^{2}x}{{\left(1-ta{n}^{2}x\right)}^2}}$

$=\frac{4tanx*\left(1-ta{n}^{2}x\right)}{1+ta{n}^{4}x-2ta{n}^{2}x-4ta{n}^{2}x}$

$=\frac{4tanx\left(1-ta{n}^{2}x\right)}{1-6ta{n}^{2}x+ta{n}^{4}x}$

= R.H.S.

39. L.H.S. = cot x cot 2x – cot 2x cot 3x – cot 3x cot x.

= cot x cot 2x – cot 3x (cot 2x + cot x)

= cot x cot 2x – (cot 2x + cot x) [cot (2x + x)]

We know that,

$cot(A+B)=\frac{co\mathrm{tA}co\mathrm{tB}-1}{co\mathrm{tA}+co\mathrm{tB}}$ we can write

$L.H.S=cotxcot2x-(cot2x+cotx)\left[\frac{cot2x·cotx-1}{cot2x+cotx}\right]$

= cot x cot 2x – cot 2x cot x + 1

= R.H.S.