# Trigonometric Equations

Solve: 2cosθ = secθ [0° ≤ θ ≤ 90°]

Here, 2cosθ = secθ or, 2cosθ = $\frac{1}{cos\theta}$
or, cos2θ = $\frac 12$
or, cosθ = ±$\frac{1}{\sqrt{2}}$
or, cosθ = cos45° (0° ≤ θ ≤ 90°)
so, θ = 45°

Solve: cosecθ = 2sinθ [0° ≤ θ ≤ 90°]

Here, cosecθ = 2sinθ or, 2sinθ = $\frac{1}{sin\theta}$
or, sin2θ = $\frac 12$
or, sinθ = ±$\frac{1}{\sqrt{2}}$
or, sinθ = sin45° (0° ≤ θ ≤ 90°)
so, θ = 45°

sin2θ - sinθ + $\frac 14$ = 0 [0° ≤ θ ≤ 90°]

Here, sin2θ - sinθ + $\frac 14$ = 0 or, (sinθ)2 - 2.sinθ.$\frac 12$ + $\left(\frac 12\right)^2$ = 0
or, $\left(\text{sinθ - }\frac 12\right)^2$ = 0

or, sinθ = $\frac 12$
or, sinθ = sin30°, sin(180 - 30°)
so, θ = 30°, 150°

Since, 0° ≤ θ ≤ 90°, θ = 30°

$\sqrt3$tanθ + 3 = 0 [0° ≤ θ ≤ 90°]

Here, $\sqrt3$tanθ = -3 or, tanθ = $\frac{-\sqrt3.\sqrt3}{\sqrt3}$
or, tanθ = -$\sqrt3$
or, tanθ = -tan60°
or, tanθ = tan(180° - 60), tan(360° - 60°)
So, θ = 120°, 300°

Since, 0° ≤ θ ≤ 90°, θ = 120°

4 - 3sec2θ = 0 [0° ≤ θ ≤ 90°]

Here, 4 - 3sec2θ = 0
or, 4 - $\frac{3}{cos^{2}θ}$ = 0
or, 4cos2θ - 3 = 0
or, cos2θ = $\frac 34$
or, cosθ = ±$\frac{\sqrt3}{2}$

Taking positive sign,
or, cosθ = cos30°, cos(360° - 30°)
So, θ = 30°, 330°
Taking negative sign,
or, cosθ = -cos30°
or, cosθ = cos(180° - 30°), cos(180° + 30°)
So, θ = 150°, 210°

Since, 0° ≤ θ ≤ 90°, θ = 30°

1 - tan2θ = -2 [0° ≤ θ ≤ 90°]

Here, 1 - tan2θ = -2 or, tan2θ = 3
or, tanθ = ±$\sqrt3$

Taking positive sign,
or, tanθ = tan60°, tan(180° + 60°)
So, θ = 60°, 240°
Taking negative sign,
or, tanθ = -tan60°
or, tanθ = tan(180° - 60°), tan(360° - 60°)
So, θ = 120°, 300°

Since, 0° ≤ θ ≤ 90°, θ = 60°

$\sqrt3$tanA + 1 = 0 [0° ≤ θ ≤ 180°]

Here, $\sqrt3$tanA + 1 = 0 or, tanA = -$\frac{1}{\sqrt3}$
or, tanA = -tan30°
or, tanA = tan(360° - 30°), tan(180° - 30°)
So, A = 330°, 150°

Since, 0° ≤ θ ≤ 180°, A = 150°

sinx - sin2x = 0 [0° ≤ x ≤ 90°]

Here, sinx - sin2x = 0 or, sinx - 2sinx.cosx = 0
or, sinx(1 - 2cosx) = 0

Either, sinx = 0
or, sinx = sin0°, sin(180° - 0°)
So, x = 0°, 180°
OR, 1 - 2cosx = 0
or, cosx = $\frac 12$
or, cosx = cos60°, cos(360° - 60°)
So, x = 60°, 300°

Since, 0° ≤ x ≤ 90°, x = 0°, 60°

sin2θ + cosθ = 0 [0° ≤ θ ≤ 90°]

Here, sin2θ + cosθ = 0 or, 2sinθ.cosθ + cosθ = 0
or, cosθ(2sinθ + 1) = 0

Either, cosθ = 0
or, cosθ = cos90°, cos(360° - 90°)
So, θ = 90°, 270°
OR, 2sinθ + 1 = 0
or, 2sinθ = -1
or, sinθ = -$\frac 12$
0r, sinθ = -sin30°
or, sinθ = sin(180° + 30°), sin(360° - 30°)
So, θ = 210°, 330°

Since, 0° ≤ θ ≤ 90°, θ = 90°

3tanθ - $\sqrt3$ = 0 [0° ≤ θ ≤ 90°]

Here, 3tanθ - $\sqrt3$ = 0 or, tanθ = $\frac{\sqrt3}{3}$
or, tanθ = $\frac{1}{\sqrt3}$
or, tanθ = tan30°, tan(180° + 30°)
So, θ = 30°, 210°

Since, 0° ≤ θ ≤ 90°, θ = 30°

3secθ = 4cosθ [0° ≤ θ ≤ 90°]

Here, 3secθ = 4cosθ or, 3 = 4cos2θ
or, cos2θ = $\frac 34$
or, cosθ = $\frac{\sqrt3}{2}$
or, cosθ = cos30°, cos(360° - 30°)
So, θ = 30°, 330°

Since, 0° ≤ θ ≤ 90°, θ = 30°

$\sqrt3$tanθ - 3 = 0 [0° ≤ θ ≤ πc]

Here, $\sqrt3$tanθ - 3 = 0 or, tanθ = $\frac{3}{\sqrt3}$
or, tanθ = $\sqrt3$
or, tanθ = tan60°, tan(180° + 60°)
So, θ = 60°, 240°

Since, 0° ≤ θ ≤ πc, θ = 60°

tanθ = cotθ [0° ≤ θ ≤ 90°]

Here, tanθ = cotθ or, tan2θ = 1
or, tanθ = tan45°, tan(180° + 45°)
So, θ = 45°, 225°

Since, 0° ≤ θ ≤ 90°, θ = 45°

tan2A - 1 = 2 [0° ≤ A ≤ 180°]

Here, tan2A - 1 = 2 or, tan2A = 3
or, tanA = ±$\sqrt3$

Taking positive,
or, tanA = tan60°, tan(180° + 60°)
So, A = 60°, 240°
Taking negative,
or, tanA = -tan60°
or, tanA = tan(180° - 60°), tan(360° - 60°)
So, A = 120°, 300°

Since, 0° ≤ A ≤ 180°, A = 60°, 120°

$\sqrt2$secθ + 2 = 0 [0° ≤ θ ≤ 360°]

Here, $\sqrt2$secθ + 2 = 0 or, secθ = -$\frac{2}{\sqrt2}$
or, secθ = -$\sqrt2$
or, cosθ = -$\frac{1}{\sqrt2}$
or, cosθ = -cos45°
or, cosθ = cos(180° - 45°), cos(180° + 45°)
So, θ = 135°, 225°

As, 0° ≤ θ ≤ 360°, θ = 135°, 225°

2sin2α - 1 = 0 [0° ≤ α ≤ 180°]

Here, 2sin2α - 1 = 0 or, sinα = ±$\frac{1}{\sqrt2}$

Taking positive,
or, sinα = sin45°, sin(180° - 45°)
So, α = 45°, 135°
Taking negative sign,
or, sinα = -sin45°
or, sinα = sin(180° + 45°), sin(360° - 45°)
So, α = 225°, 315°

Since, 0° ≤ α ≤ 180°, α = 45°, 135°

If 2sinθ - $\sqrt3$ = 0, find the acute value of θ.

Here, 2sinθ - $\sqrt3$ = 0 or, sinθ = $\frac{\sqrt3}{2}$
or, sinθ = sin60°, sin(180° - 60°)
So, θ = 60°, 120°

Hence, acute value of θ is 60°

Note: Acute value is any value less than 90°. Similarly, obtuse value is any value more than 90° and less than 180°. However, reflex value is any value more than 180° and less than 360°.

2cos2θ + $\sqrt3$cosθ = 0 [0° ≤ θ ≤ 180°]

Here, 2cos2θ + $\sqrt3$cosθ = 0 or, cosθ(2cosθ + + $\sqrt3$) = 0

Either, cosθ = 0
or, cosθ = cos90°, cos(360° - 90°)
So, θ = 90°, 270°
OR, 2cosθ + + $\sqrt3$ = 0
or, cosθ = -$\frac{\sqrt3}{2}$
or, cosθ = -cos30°
or, cosθ = cos(180° - 30°), cos(180° + 30°)
So, θ = 150°, 210°

Since, 0° ≤ θ ≤ 180°, θ = 90°, 150°

2cos2θ - 1 = 0 [0° ≤ θ ≤ 180°]

Here, 2cos2θ - 1 = 0 or, cos2θ = $\frac 12$
or, cosθ = ±$\frac{1}{\sqrt2}$

Taking positive,
or, cosθ = cos45°, cos(360° - 45°)
So, θ = 45°, 315°
Taking negative,
or, cosθ = -cos45°
or, cosθ = cos(180° + 45°), cos(180° - 45°)
So, θ = 225°, 135°

Since, 0° ≤ θ ≤ 180°, θ = 45°, 135°

$\frac{\sqrt2}{cosB}$ + 2 = 0 [0° ≤ B ≤ 180°]

Here, $\frac{\sqrt2}{cosB}$ + 2 = 0 or, $\sqrt2$ + 2cosB = 0
or, cosB = -$\frac{\sqrt2}{2}$
or, cosB = -$\frac{\sqrt2}{\sqrt2 \times \sqrt2}$
or, cosB = -$\frac{1}{\sqrt2}$
or, cosB = -cos45°
or, cosB = cos(180° - 45°), cos(180° + 45°)
So, B = 135°, 225°

Since, 0° ≤ B ≤ 360°, B = 135°

If sinθ = cosθ, find the acute angled value of 'θ'.

Here, sinθ = cosθ Squaring on both side, we get,
or, sin2θ = cos2θ
or, sin2θ = 1 - sin2θ
or, 2sin2θ = 1
or, sinθ = ±$\frac{1}{\sqrt2}$

Taking positive,
or, sinθ = sin45°, sin(180° - 45°)
So, θ = 45°, 135°
Taking negative,
or, sinθ = -sin45°
or, sinθ = sin(180° + 45°), sin(360° - 45°)
So, θ = 225°, 315°

Hence, acute angled value of 'θ' is 45°

cot2x = 3 [0° ≤ θ ≤ 180°]

Here, cot2x = 3 or, tanx = ±$\frac{1}{\sqrt3}$

Taking positive,
or, tanx = tan30°, tan(180° + 30°)
So, x = 30°, 210°
Taking negative,
tanx = -tan30°
or, tanx = tan(180° - 30°), tan(360° - 30°)
So, x = 150°, 330°

Since, 0° ≤ θ ≤ 180°, θ = 30°, 150°