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 ≤ π^c]

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 ≤ π^c, 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°