# Long Question

$\sqrt{3}$tanθ = $\sqrt{3}$secθ - 1 (0° ≤ θ ≤ 360°)

Here, $\sqrt{3}$tanθ = $\sqrt{3}$secθ - 1 or, $\sqrt{3}\frac{sin\theta}{cos\theta}$ = $\sqrt{3}\frac{1}{cos\theta}$ - 1
or, $\sqrt{3}$sinθ + cosθ = $\sqrt{3}$ --- (i)

Now, Dividend factor, d.f. = $\sqrt{(\text{Co-efficient of sinθ})^2 + (\text{Co-efficient of cosθ})^2}$ = 2

Dividing both sides of (i) by d.f., we get, $\frac{\sqrt{3}}{2}$sinθ + $\frac 12$cosθ = $\frac{\sqrt{3}}{2}$ or, sinθ.cos30° + cosθ.sin30° = $\frac{\sqrt{3}}{2}$
or, sin(θ + 30°) = sin60°, sin(180° - 60°)
so, comparing the angles, we get,
θ = 30°, 90° where 0° ≤ θ ≤ 360°

2$\sqrt3$cos2θ = sinθ [0° ≤ θ ≤ 360°]

Here, 2$\sqrt3$cos2θ = sinθ or, 2$\sqrt3$(1 - sin2θ) = sinθ
or, 2$\sqrt3$ - 2$\sqrt3$sin2θ = sinθ
or, 2$\sqrt3$sin2θ + sinθ - 2$\sqrt3$ = 0
or, 2$\sqrt3$sin2θ + 4sinθ - 3sinθ - 2$\sqrt3$ = 0
or, 2sinθ($\sqrt3$sinθ + 2) - $\sqrt3$($\sqrt3$sinθ + 2) = 0
or, ($\sqrt3$sinθ + 2)(2sinθ - $\sqrt3$) = 0

Either, $\sqrt3$sinθ + 2 = 0
or, sinθ = $\frac{2}{\sqrt3}$
It has no solution.
OR, 2sinθ - $\sqrt3$ = 0
or, sinθ = $\frac{\sqrt3}{2}$
or, sinθ = sin60°, sin(180° - 60°)
So, θ = 60°, 120°

As, 0° ≤ θ ≤ 360°, θ = 60°, 120°

sinθ + cosθ = 1 [0° ≤ θ ≤ 360°]

Here,
sinθ + cosθ = 1 ---- (i)

Now, Dividend factor,
d.f. = $\sqrt{(\text{Co-efficient of sinθ})^2 + (\text{Co-efficient of cosθ})^2}$

= $\sqrt2$

Dividing both side of (i) by d.f. we get, $\frac{1}{\sqrt2}$sinθ + $\frac{1}{\sqrt2}$cosθ = $\frac{1}{\sqrt2}$ or, sinθ.cos45° + cosθ.sin45° = sin45°
or, sin(θ + 45°) = sin45°, sin(180° - 45°), sin(360° + 45°)
or, sin(θ + 45°) = sin45°, sin135°, sin405°

Either, θ + 45° = 45°
So, θ = 0°

OR, θ + 45° = 135°
So, θ = 90°

OR, θ + 45° = 405°
So, θ = 360°

Since, 0° ≤ θ ≤ 360°, θ = 0°, 90°, 360°

$\sqrt3$sinθ + cosθ = $\sqrt2$ [0° ≤ θ ≤ 2πc]

Here,
$\sqrt3$sinθ + cosθ = $\sqrt2$ ---- (i)

Now, Dividend factor,
d.f. = $\sqrt{(\text{Co-efficient of sinθ})^2 + (\text{Co-efficient of cosθ})^2}$ = $\sqrt{\text{3 + 1}}$
= 2

Dividing both side of (i) by d.f. we get, $\frac{\sqrt3}{2}$sinθ + $\frac{1}{2}$cosθ = $\frac{\sqrt2}{2}$ or, sinθ.cos30° + cosθ.sin30° = sin45°
or, sin(θ + 30°) = sin45°, sin(180° - 45°), sin(360° + 45°)
or, sin(θ + 30°) = sin45°, sin135°, sin405°

Either, θ + 30° = 45°
So, θ = 15°

OR, θ + 30° = 135°
So, θ = 105°

OR, θ + 30° = 405°
So, θ = 375°

Since, 0° ≤ θ ≤ 360°, θ = 15°, 105°

1 + $\sqrt3$tanα - secα = 0 [0° ≤ α ≤ 360°]

Here, 1 + $\sqrt3$tanα - secα = 0 or, 1 + $\frac{\sqrt{3}sinα}{cosα}$ - $\frac{1}{cosα}$ = 0
or, cosα + $\sqrt3$sinα = 1 ---- (i)

Now, Dividend factor, d.f. = $\sqrt{(\text{Co-efficient of sinθ})^2 + (\text{Co-efficient of cosθ})^2}$ = $\sqrt{1 + 3}$
= 2

Dividing both side of (i) by d.f. we get, $\frac{1}{2}$cosα + $\frac{\sqrt3}{2}$sinα = $\frac{1}{2}$ or, sin30°.cosα + sin30°.cosα = sin30°
or, sin(30° + α) = sin30°, sin(180° - 30°), sin(360° + 30°)
or, sin(30°+ α) = sin30°, sin150°, sin405°

Either, 30°+ α = 30°
So, α = 0°

OR, 30°+ α = 135°
So, α = 120°

OR, 30°+ α = 390°
So, α = 360°

As, 0° ≤ α ≤ 360°, α = 0°, 120°, 360°

sinA + sin3A = sin2A [0° ≤ A ≤ 360°]

Here, sinA + sin3A = sin2A or, 2sin$\frac{\text{A + 3A}}{2}$.cos$\frac{\text{A - 3A}}{2}$ = sin2A
or, 2sin2A.cos(-A) = sin2A
or, 2sin2A.cosA - sin2A = 0 [cos(-θ) = cosθ]
or, sin2A(2cosA - 1) = 0

Either, sin2A = 0 or, 2sinA.cosA = 0
Again,
Either, sinA = 0
or, sinA = sin0°, sin(180° - 0°), sin(360° + 0°)
So, A = 0°, 180°, 360° ---- (A)
OR, cosA = 0
or, cosA = cos90°, cos(360 - 90)
So, A = 90°, 270° ---- (B)
OR, 2cosA - 1 = 0 or, cosA = $\frac 12$
or, cosA = cos60°, cos(360° - 60°)
So, A = 60°, 300° ---- (C)

As, 0° ≤ A ≤ 360°, from (A), (B)and (C) A = 0°, 60°, 90°, 180°, 270° 300°, 360°

sinA = $\sqrt3$(1 - cosA) [0° ≤ A ≤ 360°]

Here,
sinA + $\sqrt3$cosA = $\sqrt3$ ---- (i)

Now, Dividend factor, d.f. = $\sqrt{(\text{Co-efficient of sinθ})^2 + (\text{Co-efficient of cosθ})^2}$ = $\sqrt{1 + 3}$
= 2

Dividing both side of (i) by d.f. we get, $\frac 12$sinA + $\frac{\sqrt3}{2}$cosA = $\frac{\sqrt3}{2}$ or, sinA.cos60° + cosA.sin60° = sin60°
or, sin(A + 60°) = sin60°, sin(180° - 60°), sin(360° + 60°)

Either, A + 60° = 60°
So, A = 0°

OR, A + 60° = 120°
So, A = 60°

OR, A + 60° = 360° + 60°
So, A = 360°

As, 0° ≤ A ≤ 360°, A = 0°, 60°, 360°

$\sqrt3$sinθ - cos θ = $\sqrt2$ [0° ≤ θ ≤ 360°]

Here,
$\sqrt3$sinθ - cos θ = $\sqrt2$ ---- (i)

Now, Dividend factor, d.f. = $\sqrt{(\text{Co-efficient of sinθ})^2 + (\text{Co-efficient of cosθ})^2}$ = $\sqrt{1 + 3}$
= 2

Dividing both side of (i) by d.f. we get, $\frac{\sqrt3}{2}$sinθ - $\frac{1}{2}$cosθ = $\frac{\sqrt2}{2}$ or, sinθ.cos30° - cosθ.sin30° = sin45°
or, sin(θ - 30°) = sin45°, sin(180° - 45°), sin(360° + 45°)

Either, θ - 30° = 45°
So, θ = 75°

OR, θ - 30° = 135°
So, θ = 165°

OR, θ - 30° = 405°
So, θ = 435°

Since, 0° ≤ θ ≤ 360°, θ = 75°, 165°

$\sqrt3$sinx + cosx = 1 [0° ≤ x ≤ 2πc]

Here,
$\sqrt3$sinx + cosx = 1 ---- (i)

Now, Dividend factor, d.f. = $\sqrt{(\text{Co-efficient of sinθ})^2 + (\text{Co-efficient of cosθ})^2}$ = $\sqrt{1 + 3}$
= 2

Dividing both side of (i) by d.f. we get, $\frac{\sqrt3}{2}$sinx + $\frac{1}{2}$cosx = $\frac{1}{2}$ or, sinx.cos30° + cosx.sin30° = sin30°
or, sin(x + 30°) = sin30°, sin(180° - 30°), sin(360° + 30°)

Either, x + 30° = 30°
So, x = 0°

OR, x + 30° = 150°
So, x = 120°

OR, x + 30° = 390°
So, x = 360°

Since, 0° ≤ x ≤ 360°, x = 0°, 120°, 360°

cosθ + cos2θ + cos3θ = 0 [0° ≤ θ ≤ 180°]

Here, cosθ + cos2θ + cos3θ = 0 or, 2cos$\frac{\text{θ + 30}}{2}$cos$\frac{\text{θ - 30}}{2}$ + cos2θ = 0
or, 2cos2θ.cosθ + cos2θ = 0 [cos(-θ) = cosθ]
or, cos2θ(2cosθ + 1) = 0

Either, cos2θ = 0
or, 2cos2θ - 1 = 0
or, cosθ = ±$\frac{1}{\sqrt2}$
Taking positive,
cosθ = cos45°, cos(360° - 45°)
So, θ = 45°, 315° ---- (A) Taking negative,
cosθ = -cos45°
cosθ = cos(180° - 45°), cos(180° + 45°)
So, θ = 135°, 225° ---- (B)
OR, cosθ = -$\frac 12$
or, cosθ = cos60°
or, cosθ = cos(180° - 60°), cos(180° + 60°)
So, θ = 120°, 240° ---- (C)

Since, 0° ≤ θ ≤ 180°, from (A), (B) and (C) θ = 45°, 120°, 135°

(1 - $\sqrt3$)tanθ + 1 + $\sqrt3$ = $\sqrt3$sec2θ [0° ≤ θ ≤ 360°]

Here, (1 - $\sqrt3$)tanθ + 1 + $\sqrt3$ = $\sqrt3$sec2θ or, (1 - $\sqrt3$)tanθ + 1 + $\sqrt3$ = $\sqrt3$(1 + tan2θ)
or, (1 - $\sqrt3$)tanθ + 1 + $\sqrt3$ = $\sqrt3$ + $\sqrt3$tan2θ
or, $\sqrt3$tan2θ - (1 - $\sqrt3$)tanθ - 1 = 0
or, $\sqrt3$tan2θ - tanθ + $\sqrt3$tanθ - 1 = 0
or, tanθ($\sqrt3$tanθ - 1) + 1($\sqrt3$tanθ - 1) = 0
or, (tanθ + 1)($\sqrt3$tanθ - 1) = 0

Either, tanθ = -1
or, tanθ = -tan45°
or, tanθ = tan(180° - 45°), tan(360° - 45°)
So, θ = 135°, 315°
OR, tanθ = $\frac{1}{\sqrt3}$
or, tanθ = tan30°
or, tanθ = tan30°, tan(180° + 30°)
So, θ = 30°, 210°

As, 0° ≤ θ ≤ 360°, θ = 30°, 135°, 210°, 315°

2$\sqrt3$sin2θ = cosθ [0° ≤ θ ≤ 360°]

Here, 2$\sqrt3$sin2θ = cosθ or, 2$\sqrt3$(1 - cos2θ) = cosθ
or, 2$\sqrt3$ - 2$\sqrt3$cos2θ = cosθ
or, 2$\sqrt3$cos2θ + cosθ - 2$\sqrt3$ = 0
or, 2$\sqrt3$cos2θ + 4cosθ - 3cosθ - 2$\sqrt3$ = 0
or, 2cosθ($\sqrt3$cosθ + 2) - $\sqrt3$($\sqrt3$cosθ + 2) = 0
or, (2cosθ - $\sqrt3$)($\sqrt3$cosθ + 2) = 0

Either, 2cosθ - $\sqrt3$ = 0
or, cosθ = $\frac{\sqrt3}{2}$
or, cosθ = cos30°, cos(360° - 30°)
So, θ = 30°, 330°
OR, cosθ = -$\frac{2}{\sqrt3}$
It has not solution.

As, 0° ≤ θ ≤ 360°, θ = 30°, 330°

3sin2x + 4cosx = 4 [0° ≤ θ ≤ 2πc]

Here, 3sin2x + 4cosx = 4 or, 3(1 - cos2x) + 4cosx - 4 = 0
or, 3 - 32 + 4cosx - 4 = 0
or, 3cos2x - 4cosx + 1 = 0
or, 3cos2x - (3 + 1)cosx + 1 = 0
or, 3cos2x - 3cosx - cosx + 1 = 0
or, 3cosx(cosx - 1) - 1(cosx - 1) = 0
or, (3cosx - 1)(cosx - 1) = 0

Either, 3cosx - 1 = 0
or, cosx = $\frac 13$
So, x = cos-1$\left(\frac 13\right)$
OR, cosx - 1 = 0
or, cosx = 1
or, cosx = cos0°, cos(360° - 0°)
So, x = 0°, 360°

Hence, 0° ≤ θ ≤ 2πc, x = 0°, cos-1$\left(\frac 13\right)$, 360°

$\sqrt3$cosx = $\sqrt3$ - sinx [0° ≤ θ ≤ 360°]

Here,
sinx + $\sqrt3$cosx = $\sqrt3$

Now, Dividend factor, d.f. = $\sqrt{(\text{Co-efficient of sinθ})^2 + (\text{Co-efficient of cosθ})^2}$ = $\sqrt{1 + 3}$
= 2

Dividing both side of (i) by d.f. we get, $\frac{1}{2}$sinx + $\frac{\sqrt3}{2}$cosx = $\frac{\sqrt3}{2}$ or, sinx.cos60° + cosx.sin60° = sin60°
or, sin(x + 60°) = sin60°, sin(180° - 60°), sin(360° + 60°)

Either, x + 60° = 60°
So, x = 0°

OR, x + 60° = 120°
So, x = 60°

OR, x + 60° = 360° + 60°
So, x = 360°

Since, 0° ≤ x ≤ 360°, x = 0°, 60°, 360°

2sin2θ - cosθ = 1 [0° ≤ θ ≤ 2πc]

Here, 2sin2θ - cosθ = 1 or, 2(1 - cos2θ) - cosθ - 1 = 0
or, 2 - 2cos2θ - cosθ - 1 = 0
or, 2cos2θ + cosθ - 1 = 0
or, 2cos2θ + 2cosθ - cosθ - 1 = 0
or, 2cosθ(cosθ + 1) - 1(cosθ + 1) = 0
or, (2cosθ - 1)(cosθ + 1) = 0

Either, 2cosθ - 1 = 0
or, cosθ = $\frac 12$
or, cosθ = cos60°, cos(360° - 60°)
So, θ = 60°, 300°
OR, cosθ = -1
or, cosθ = -cos0°
or, cosθ = cos(180° - 0°)
So, θ = 180°

Hence, 0° ≤ θ ≤ 2πc, θ = 60°, 180°, 300°

cos2θ - sinθ - $\frac 14$ = 0 [0° ≤ θ ≤ 360°]

Here, cos2θ - sinθ - $\frac 14$ = 0 or, 1 - sin2θ - sinθ - $\frac 14$ = 0
or, sin2θ + sinθ + $\frac 14$ = 1
or, (sinθ)2 + 2.sinθ.$\frac 12$ + $\left(\frac 12\right)^2$ = 1
or, $\left(sinθ \space + \space \frac 12\right)^2$ = 1
or, sinθ = 1 - $\frac 12$
or, sinθ = $\frac 12$
or, sinθ = sin30°, sin(180° - 30°)
So, θ = 30°, 150°

As, 0° ≤ θ ≤ 360°, θ = 30°, 150°

6sin2θ + cosθ = 5 [0° ≤ θ ≤ 2πc]

Here, 6sin2θ + cosθ = 5 or, 6(1 - cos2θ) + cosθ = 5
or, 6 - 6cos2θ + cosθ = 5
or, 6cos2θ - cosθ - 1 = 0
or, 6cos2θ - (3 - 2)cosθ - 1 = 0
or, 6cos2θ - 3cosθ + 2cosθ - 1 = 0
or, 3cosθ(2cosθ - 1) + 1(2cosθ - 1) = 0
or, (3cosθ + 1)(2cosθ - 1) = 0

Either, 3cosθ + 1 = 0
or, cosθ = -$\frac 13$
So, θ = cos-1$\left(\frac{-1}{3}\right)$
OR, cosθ = $\frac 12$
or, cosθ = cos60°, cos(360° - 60°)
So, θ = 60°, 300°

As, 0° ≤ θ ≤ 2πc, θ = 60°, cos-1$\left(\frac{-1}{3}\right)$, 300°