Ans.

(i)

Sol. The RHS of the given transformation formulae are not defined for x = (2n +1) π

Therefore in case they satisfy the given equation they will disappear from the solution set which can be actually observed as

+ = –1

⇒ t = –1

⇒ tan = –1

⇒ = n π –

⇒ x = 2n π –

This will never yield (2n + 1) π for any integer though the various (2n + 1) π satisfies the given trigonometric equation.

(ii)

Sol. On squaring, we get sin 2x = 0

⇒ 2x = n π

⇒ x = n π /2 (*)

But this includes the solution of the equation sin x + cos x

= –1 also.

Therefore we will find integer n given that a solution from (*) does not satisfy sin x + cos x = –1.

Indeed will be solution of sin x + cos x = +1

If sin + cos = +1

We easily observe that when

n = 4m, 4m + 1, the values satisfies, sin x + cos x = 1 and when n is 4m + 2 or 4m + 3 then

satisfies sin x + cos x = –1

Note : The equation sin x + cos x = 1 can easily be solved by writing it as sin = 1

⇒ x + = 2n π + (–1) n

Take two cases n odd, n even.

The above has been described for cases when squaring is indispensable.

(iii)

Sol. The equation has a meaning if sin x ≤ which is always true. Any x for which sin x < cannot be solution since

> 0 for all x.

On squaring the equation and solving,

we get sin x = – , sin x =

But sin x = – is not possible.

⇒ sin x =

When the correct choice is

(iv)

Sol. The equation has a meaning if 2x ≠ n π , 3x = n π , 2x

≠ (2n +1)

For these values the equation can be written as

3 = +

⇒ =

⇒ 3 sin x cos 2x = sin 3x

⇒ sin x(3 – 4 sin 2 x –3 cos 2x) = 0

⇒ sin x sin 2 x = 0

⇒ sin x = 0 ⇒ x = n π

But for x = n π , the equation has no meaning

⇒ The given equation has no solution

(v)

Sol. If cos x ≥ 0 then the equation is equivalent to

cos x = cos x –2 sin x

⇒ sin x = 0 ⇒ x = n π

But in order to have cos x > 0, n should be even

Again if cos x < 0, then the equation is equivalent to tanx = 1

⇒ x = n π +

But in order to have cos x < 0 we must choose n odd

Thus the correct solution set is n π (n even) or n π + ,

(n odd)