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Question 1:

Find the general solutions of the following equations:
(i)

(ii)

(iii)

(iv)

(v)

(vi)

We have:
(i)
The value of x satisfying  is $\frac{\mathrm{\pi }}{6}$.
∴
⇒
⇒

(ii)
The value of x satisfying  is $\frac{7\mathrm{\pi }}{6}$.

⇒
⇒

(iii) (or)
The value of x satisfying  is $-\frac{\mathrm{\pi }}{4}$.
∴
⇒
⇒ ,
⇒

(iv) (or)
The value of x satisfying  is $\frac{\mathrm{\pi }}{4}$.
∴
⇒
⇒

(v)
The value of x satisfying  is $-\frac{\mathrm{\pi }}{6}$.

⇒

(vi)
⇒   (or)
The value of x satisfying  is $\frac{\mathrm{\pi }}{6}$.
∴

,

Question 2:

Find the general solutions of the following equations:
(i)
(ii)
(iii)
(iv)
(v)
(vi)
(vii)
(viii)
(ix)
(x)
(xi)
(xii)

We have:

(i)
⇒
⇒
⇒

(ii)
⇒
⇒
⇒

(iii)
⇒
⇒
⇒ or
⇒ or
⇒ , or ,
⇒ or ,

(iv)

⇒
⇒
On taking positive sign, we have:

⇒

⇒
Now, on taking negative sign, we have:

⇒
⇒

(v)

(vi)

(vii)

(viii)

(ix)

(x)

On taking positive sign, we have:

On taking negative sign, we have:

(xi)

or
Now,

(xii)

On taking positive sign, we have:

On taking negative sign, we have:

Question 3:

Solve the following equations:
(i)
(ii)
(iii)
(iv)
(v)
(vi)
(vii)

(i)

or
or
$\mathrm{cos}x=-\frac{3}{2}$ is not possible.

(ii)

or,
is not possible.

(iii)

$⇒$    or

is not possible.

(iv) $4{\mathrm{sin}}^{2}x-8\mathrm{cos}x+1=0$

or
Now,
(It is not possible.)

(v)

or
Now,

And,

(vi)
Now,

(vii)

On taking positive sign, we have:

On taking negative sign, we have:

Question 4:

Solve the following equations:
(i)
(ii)
(iii)
(iv)
(v)
(vi)
(vii)
(viii)
(ix)

(i)
Now,

or,
or
,   or
or

(ii)

or
or
or
or

(iii)

or
or
or
$⇒$    or

(v)

or
or
$⇒$   or
or
$⇒$  or        (Taking negative sign will give absurd result.)
or

(vi)

or
or
$⇒$  or

(vii)

or
or
or,
or     or

(viii)

or
$⇒$   or
,   or
or

(ix)

or
,   or
or

Question 5:

Solve the following equations:
(i)
(ii)
(iii)

(i) We have:
Now,

or

Now,

And,

∴   or
Here,

(ii) Given:

Now,

or
Now,

And,

∴  or

(iii) Given:
Now,

or,  or,
And,
Or,
And,

∴  or

Question 6:

Solve the following equations:
(i)
(ii)
(iii)
(iv)

(i) Given:
...(i)
The equation is of the form , where  and .
Let: and
Now, and
On putting  and in equation (i), we get:

(ii) Given: ...(ii)
The equation is of the form of , where  and .
Let:  and
Now, and
On putting and  in equation (ii), we get:

On taking positive sign, we get:

Now, on taking negative sign of the equation, we get:

(iii) Given:      ...(iii)
The equation is of the form , where  and .
Let:  and
Now, and
On putting  and  in equation (iii), we get:

On taking positive sign, we get:

On taking negative sign, we get:

(iv) Given:

The equation is of the form , where  and .
Let: and
Now, and
On putting  and   in equation (iv), we get:

On taking positive sign, we get:

On taking negative sign, we get:

Question 7:

Solve the following equations:
(i) $\mathrm{cot}x+\mathrm{tan}x=2$                                                                                       [NCERT EXEMPLAR]
(ii)                                                                      [NCERT EXEMPLAR]
(iii)                                                                [NCERT EXEMPLAR]
(iv)                                                                          [NCERT EXEMPLAR]
(v) $\mathrm{sin}x-3\mathrm{sin}2x+\mathrm{sin}3x=\mathrm{cos}x-3\mathrm{cos}2x+\mathrm{cos}3x$                                           [NCERT EXEMPLAR]
(vi) 4sinx cosx + 2 sin x + 2 cosx + 1 = 0
(vii) cosx + sin x = cos 2x + sin 2x
(viii) sin x tan x – 1 = tan x – sin x
(ix) 3tanx + cot x = 5 cosec x

(i)
$\mathrm{cot}x+\mathrm{tan}x=2\phantom{\rule{0ex}{0ex}}⇒\frac{1}{\mathrm{tan}x}+\mathrm{tan}x=2\phantom{\rule{0ex}{0ex}}⇒{\mathrm{tan}}^{2}x+1=2\mathrm{tan}x\phantom{\rule{0ex}{0ex}}⇒{\mathrm{tan}}^{2}x-2\mathrm{tan}x+1=0\phantom{\rule{0ex}{0ex}}⇒{\left(\mathrm{tan}x-1\right)}^{2}=0$

(ii)
$2{\mathrm{sin}}^{2}x=3\mathrm{cos}x\phantom{\rule{0ex}{0ex}}⇒2\left(1-{\mathrm{cos}}^{2}x\right)=3\mathrm{cos}x\phantom{\rule{0ex}{0ex}}⇒2{\mathrm{cos}}^{2}x+3\mathrm{cos}x-2=0\phantom{\rule{0ex}{0ex}}⇒\left(2\mathrm{cos}x-1\right)\left(\mathrm{cos}x+2\right)=0$

But, $\mathrm{cos}x=-2$ is not possible.         $\left(-1\le \mathrm{cos}x\le 1\right)$
$\therefore \mathrm{cos}x=\frac{1}{2}=\mathrm{cos}\frac{\mathrm{\pi }}{3}\phantom{\rule{0ex}{0ex}}⇒x=2n\mathrm{\pi }±\frac{\mathrm{\pi }}{3},n\in \mathbf{Z}$
Putting n = 0 and n = 1, we get

(iii)

Putting n = 0 and m = 0, we get

(iv)

(v)
$\mathrm{sin}x-3\mathrm{sin}2x+\mathrm{sin}3x=\mathrm{cos}x-3\mathrm{cos}2x+\mathrm{cos}3x\phantom{\rule{0ex}{0ex}}⇒2\mathrm{sin}2x\mathrm{cos}x-3\mathrm{sin}2x=2\mathrm{cos}2x\mathrm{cos}x-3\mathrm{cos}2x\phantom{\rule{0ex}{0ex}}⇒\mathrm{sin}2x\left(2\mathrm{cos}x-3\right)=\mathrm{cos}2x\left(2\mathrm{cos}x-3\right)\phantom{\rule{0ex}{0ex}}⇒\left(\mathrm{sin}2x-\mathrm{cos}2x\right)\left(2\mathrm{cos}x-3\right)=0$

But, $\mathrm{cos}x=\frac{3}{2}$ is not possible.        $\left(-1\le \mathrm{cos}x\le 1\right)$

(vi)

(vii)

(viii)

(ix)

Question 8:

Solve the following equations:
3 – 2 cos x – 4 sin x – cos 2x + sin 2x = 0

Question 9:

Solve the following equations:
3sin2x – 5 sin x cos x + 8 cos2 x = 2

Question 10:

Solve the following equations:
${2}^{{\mathrm{sin}}^{2}x}+{2}^{{\mathrm{cos}}^{2}x}=2\sqrt{2}$

Question 13:

If secx cos5x + 1 = 0, where $0, find the value of x.

The given equation is secx cos5x + 1 = 0.
Now,

Putting n = 0 and n = 1, we get

Also, putting m = 0, we get

Hence, the values of x are $\frac{\mathrm{\pi }}{6},\frac{\mathrm{\pi }}{4}$ and $\frac{\mathrm{\pi }}{2}$.

Question 1:

Write the number of solutions of the equation tan x + sec x = 2 cos x in the interval [0, 2π].

Given:
tanx + secx = 2 cosx

Now,

And,

Hence, the given equation has two solutions in .

Question 2:

Write the number of solutions of the equation .

We have:
...(i)
The equation is of the form , where  and .
Now,
Let:
and
Thus, we have:
and
By putting  and in equation (i), we get:

The solution is not possible.
Hence, the given equation has no solution.

Question 3:

Write the general solutions of tan2 2x = 1.

Given:

Hence, the general solution of the equation is

Question 4:

Write the set of values of a for which the equation has no solution.

Given:

If  or  , then the equation will possess a solution.
For no solution, .

Question 5:

If cos x = k has exactly one solution in [0, 2π], then write the values(s) of k.

Given:
If , then

Now,   for

If  then

Now,  for

If  then

Now,
when
And,
when

Clearly, we can see that for  has exactly one solution.
∴

Question 6:

Write the number of points of intersection of the curves $2y=1$ and .

Given curves: and
Now,

Also,

For the other value of n,  the value of x will not satisfy the given condition.

Hence, the number of points of intersection of the curves is two, i.e., .

Question 7:

Write the values of x in [0, π] for which and cos 2x are in A.P.

(i)

This equation is of the form , where  and .
Now,
Let:
and
Thus, we have:
and
On putting  and  in equation (1), we get:

For n = 0, the values of x are and for n = 1, the values of x are .

For the other value of n, the given condition is not true, i.e., [0, π].

Question 8:

Write the number of points of intersection of the curves .

Given:
and
Now,

Also,

The value of $\mathrm{sine}$ function lies between $-$1 and 1. Therefore, the two curves will not intersect at any point.

Hence, the number of points of intersection of the curves is 0.

Question 9:

Write the solution set of the equation in the interval [0, 2π].

Given:
Now,  or
or
is not possible.
Thus, we have:

By putting n = 0 and n = 1 in the above equation, we get:

or  in the interval
For the other value of n, x will not satisfy the given condition.
∴  and

Question 10:

Write the number of values of x in [0, 2π] that satisfy the equation .

Given equation:
Now,

Here, is not possible.
Or,

Taking positive sign,

Taking negative sign,

and   will  satisfy the given condition, i.e., x in [0, 2π].

Hence, two values will satisfy the given equation.

Question 11:

If $3\mathrm{tan}\left(x-15°\right)=\mathrm{tan}\left(x+15°\right)$, $0, find $\theta$.

Given: $3\mathrm{tan}\left(x-15°\right)=\mathrm{tan}\left(x+15°\right)$

$⇒\frac{\mathrm{tan}\left(x+15°\right)}{\mathrm{tan}\left(x-15°\right)}=3$

Applying componendo and dividendo, we have

$\frac{\mathrm{tan}\left(x+15°\right)+\mathrm{tan}\left(x-15°\right)}{\mathrm{tan}\left(x+15°\right)-\mathrm{tan}\left(x-15°\right)}=\frac{3+1}{3-1}\phantom{\rule{0ex}{0ex}}⇒\frac{\frac{\mathrm{sin}\left(x+15°\right)}{\mathrm{cos}\left(x+15°\right)}+\frac{\mathrm{sin}\left(x-15°\right)}{\mathrm{cos}\left(x-15°\right)}}{\frac{\mathrm{sin}\left(x+15°\right)}{\mathrm{cos}\left(x+15°\right)}-\frac{\mathrm{sin}\left(x-15°\right)}{\mathrm{cos}\left(x-15°\right)}}=\frac{4}{2}\phantom{\rule{0ex}{0ex}}⇒\frac{\mathrm{sin}\left(x+15°\right)\mathrm{cos}\left(x-15°\right)+\mathrm{cos}\left(x+15°\right)\mathrm{sin}\left(x-15°\right)}{\mathrm{sin}\left(x+15°\right)\mathrm{cos}\left(x-15°\right)-\mathrm{cos}\left(x+15°\right)\mathrm{sin}\left(x-15°\right)}=2\phantom{\rule{0ex}{0ex}}⇒\frac{\mathrm{sin}\left(x+15°+x-15°\right)}{\mathrm{sin}\left(x+15°-x+15°\right)}=2$

Question 12:

If $2{\mathrm{sin}}^{2}x=3\mathrm{cos}x$, where $0\le x\le 2\mathrm{\pi }$, then find the value of x.

The given equation is $2{\mathrm{sin}}^{2}x=3\mathrm{cos}x$.
Now,
$2{\mathrm{sin}}^{2}x=3\mathrm{cos}x\phantom{\rule{0ex}{0ex}}⇒2\left(1-{\mathrm{cos}}^{2}x\right)=3\mathrm{cos}x\phantom{\rule{0ex}{0ex}}⇒2{\mathrm{cos}}^{2}x+3\mathrm{cos}x-2=0\phantom{\rule{0ex}{0ex}}⇒\left(2\mathrm{cos}x-1\right)\left(\mathrm{cos}x+2\right)=0$

But, $\mathrm{cos}x=-2$ is not possible.         $\left(-1\le \mathrm{cos}x\le 1\right)$

Putting n = 0 and n = 1, we get

Question 1:

The smallest value of x satisfying the equation is
(a) $2\pi /3$
(b) $\pi /3$
(c) $\pi /6$
(d) $\pi /12$

(c) $\pi /6$

Given:

To obtain the smallest value of x, we will put  in the above equation.
Thus, we have:

Hence, the smallest value of x is $\frac{\mathrm{\pi }}{6}$.

Question 2:

If
(a) $\pi /3$
(b) $2\pi /3$
(c) $4\pi /6$
(d) $5\pi /12$

(a) $\pi /3$
Given:       ...(i)
This equation is of the form , where  and .
Let:

Now,

And,

On putting  in equation (i), we get:

For , .
∴

Question 3:

If , then the values of θ form a series in
(a) AP
(b) GP
(c) HP
(d) none of these

(a) AP
Given:

Now,

Now, on putting the value of $n$, we get:
a1

= a2

= a3

= a4
And so on.
Also,

And so on.
Thus, $x$ forms a series in AP.

Question 4:

If a is any real number, the number of roots of in the first quadrant is (are).
(a) 2
(b) 0
(c) 1
(d) none of these

(c) 1

Given:

If , then the equation becomes

There are two roots of the given equation, but we need to find the number of roots in the first quadrant.
There is exactly one root of the equation, that is, .

Question 5:

The general solution of the equation  is
(a)

(b)

(c) ​

(d) none of these

(c)
Given:

Question 6:

A solution of the equation , lies in the interval
(a)
(b)
(c)
(d)

(d)

Given:

or
or
is not possible.

∴

The values of $x$ lies in the third and fourth quadrants.
Hence, $x$ lies in .

Question 7:

The number of solution in [0, π/2] of the equation is
(a) 5
(b) 7
(c) 6
(d) none of these

(c) 6

Given:

or

or

Now,

Or,

And,

Hence, there are six solutions.

Question 8:

The general value of x satisfying the equation is given by
(a)

(b)

(c)

(d)

(b)

Given:
...(i)
This equation is of the form , where  and .
Let:
and
Now,
and
On putting and  in equation (i),  we get:

Question 9:

The smallest positive angle which satisfies the equation ​ is
(a) $\frac{5\pi }{6}$

(b) $\frac{2\pi }{3}$

(c) $\frac{\pi }{3}$

(d) $\frac{\pi }{6}$

(a) $\frac{5\pi }{6}$
Given:

or,

∴         or,   is not possible.

For n = 0, the value of .
Hence, the smallest positive angle is $\frac{5\mathrm{\pi }}{6}$.

Question 10:

If , then the values of x are
(a)

(b)

(c)

(d)

(c)

Given:

Question 11:

If , then, x is equal to
(a)

(b)

(c)

(d) none of these.

(b)
Given equation:

or
or
Now,

And,

∴

Question 12:

A value of x satisfying  is
(a) $\frac{5\pi }{3}$

(b) $\frac{4\pi }{3}$

(c) $\frac{2\pi }{3}$

(d) $\frac{\pi }{3}$

(d) $\frac{\pi }{3}$
Given equation:
...(i)
Thus, the equation is of the form , where  and .
Let:
and
and
and
On putting  and  in equation (i), we get:

Question 13:

In (0, π), the number of solutions of the equation ​ is
(a) 7
(b) 5
(c) 4
(d) 2.

(d) 2
Given equation:

Now,

, which is not possible, as it is not in the interval .

Hence, the number of solutions of the given equation is 2.

Question 14:

The number of values of ​x in [0, 2π] that satisfy the equation
(a) 1
(b) 2
(c) 3
(d) 4

(b) 2

or,
or
Here,  is not possible.
∴

Now for = 0 and 1, the values of  .
Hence, there are two solutions in .

Question 15:

If , then x =
(a) 0
(b)
(c) 1
(d) none of these

(d) none of these
Given equation:
Let :

Now,

∴

And,

Taking log on both sides, we get:

Question 16:

The equation has .... solution.
(a) finite
(b) infinite
(c) one
(d) no

(d) no
Given equation:
...(i)
Thus, the equation is of the form , where  and .
Let:
and
Now,

Aso,

On putting   and in equation (i), we get:

From here, we cannot find the value of $\theta$.

Question 17:

If , then general value of ​x is
(a)

(b)

(c)

(d)

(d)
Given equation:
...(i)
This is of the form , where  and .
Let:
and  .
Now,

And,

Putting  and  in equation (i), we get:

Question 18:

General solution of  is
(a)

(b)

(c)

(d) ​

(c)
Given:

Question 19:

The solution of the equation  lies in the interval
(a)
(b)
(c)
(d)

Given equation:

or
or
Now,  is not possible.
And,

For n = 0, , for = 1, ​ and so on.
Hence, $\frac{3\mathrm{\pi }}{2}$ lies in the interval .

Question 20:

If ​  and $0 then the solutions are

(a)

(b)

(c) ​

(d)

(b) $x=\frac{2\mathrm{\pi }}{3},\frac{4\mathrm{\pi }}{3}$

Given equation:

Or,

So, both  lie in .

Question 21:

The number of values of x in the interval [0, 5 π] satisfying the equation is
(a) 0
(b) 5
(c) 6
(d) 10

(c) 6
Given:

or

Now, is not possible, as the value of  lies between $-$1 and 1.
$⇒$
Also, sin x is positive only in first two quadrants. Therefore, sin x is positive twice in the interval $\left[0,\pi \right]$.
Hence, it is positive six times in the interval $\left[0,5\pi \right]$, viz

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