What is the general solution for 2sin(θ)-tan^2(θ)=0 ?

The general solution for 2sin(θ)-tan^2(θ)=0 is θ=2pin,θ=pi+2pin,θ=0.89590…+2pin,θ=pi-0.89590…+2pin

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

2sin(θ)-tan^2(θ)=0

Solution

2 sin (θ) - tan^{2} (θ) = 0

Solution

θ = 2 πn, θ = π + 2 πn, θ = 0.89590 \dots + 2 πn, θ = π - 0.89590 \dots + 2 πn

Degrees

θ = 0^{\circ} + 36 0^{\circ} n, θ = 18 0^{\circ} + 36 0^{\circ} n, θ = 51.33171 \dots^{\circ} + 36 0^{\circ} n, θ = 128.66828 \dots^{\circ} + 36 0^{\circ} n

Solution steps

2 sin (θ) - tan^{2} (θ) = 0

Express with sin, cos

- tan^{2} (θ) + 2 sin (θ)

Use the basic trigonometric identity:

tan (x) = \frac{sin ( x )}{cos ( x )}

= - (\frac{sin ( θ )}{cos ( θ )})^{2} + 2 sin (θ)

Simplify

- (\frac{sin ( θ )}{cos ( θ )})^{2} + 2 sin (θ) : \frac{- sin ^{2} ( θ ) + 2 cos ^{2} ( θ ) sin ( θ )}{cos ^{2} ( θ )}

- (\frac{sin ( θ )}{cos ( θ )})^{2} + 2 sin (θ)

Apply exponent rule:

(\frac{a}{b})^{c} = \frac{a ^{c}}{b ^{c}}

= - \frac{sin ^{2} ( θ )}{cos ^{2} ( θ )} + 2 sin (θ)

Convert element to fraction:

2 sin (θ) = \frac{2 sin ( θ ) cos ^{2} ( θ )}{cos ^{2} ( θ )}

= - \frac{sin ^{2} ( θ )}{cos ^{2} ( θ )} + \frac{2 sin ( θ ) cos ^{2} ( θ )}{cos ^{2} ( θ )}

Since the denominators are equal, combine the fractions:

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

= \frac{- sin ^{2} ( θ ) + 2 sin ( θ ) cos ^{2} ( θ )}{cos ^{2} ( θ )}

= \frac{- sin ^{2} ( θ ) + 2 cos ^{2} ( θ ) sin ( θ )}{cos ^{2} ( θ )}

\frac{- sin ^{2} ( θ ) + 2 cos ^{2} ( θ ) sin ( θ )}{cos ^{2} ( θ )} = 0

\frac{f ( x )}{g ( x )} = 0 \Rightarrow f (x) = 0

- sin^{2} (θ) + 2 cos^{2} (θ) sin (θ) = 0

Factor

- sin^{2} (θ) + 2 cos^{2} (θ) sin (θ) : sin (θ) (- sin (θ) + 2 cos^{2} (θ))

- sin^{2} (θ) + 2 cos^{2} (θ) sin (θ)

Apply exponent rule:

a^{b + c} = a^{b} a^{c}

sin^{2} (θ) = sin (θ) sin (θ)

= - sin (θ) sin (θ) + 2 sin (θ) cos^{2} (θ)

Factor out common term

sin (θ)

= sin (θ) (- sin (θ) + 2 cos^{2} (θ))

sin (θ) (- sin (θ) + 2 cos^{2} (θ)) = 0

Solving each part separately

sin (θ) = 0 or - sin (θ) + 2 cos^{2} (θ) = 0

sin (θ) = 0 : θ = 2 πn, θ = π + 2 πn

sin (θ) = 0

General solutions for

sin (θ) = 0

sin (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

θ = 0 + 2 πn, θ = π + 2 πn

θ = 0 + 2 πn, θ = π + 2 πn

Solve

θ = 0 + 2 πn : θ = 2 πn

θ = 0 + 2 πn

0 + 2 πn = 2 πn

θ = 2 πn

θ = 2 πn, θ = π + 2 πn

- sin (θ) + 2 cos^{2} (θ) = 0 : θ = arcsin (\frac{17 - 1}{4}) + 2 πn, θ = π - arcsin (\frac{17 - 1}{4}) + 2 πn

- sin (θ) + 2 cos^{2} (θ) = 0

Rewrite using trig identities

- sin (θ) + 2 cos^{2} (θ)

Use the Pythagorean identity:

cos^{2} (x) + sin^{2} (x) = 1

cos^{2} (x) = 1 - sin^{2} (x)

= - sin (θ) + 2 (1 - sin^{2} (θ))

- sin (θ) + (1 - sin^{2} (θ)) \cdot 2 = 0

Solve by substitution

- sin (θ) + (1 - sin^{2} (θ)) \cdot 2 = 0

Let:

sin (θ) = u

- u + (1 - u^{2}) \cdot 2 = 0

- u + (1 - u^{2}) \cdot 2 = 0 : u = - \frac{1 + 17}{4}, u = \frac{17 - 1}{4}

- u + (1 - u^{2}) \cdot 2 = 0

Expand

- u + (1 - u^{2}) \cdot 2 : - u + 2 - 2 u^{2}

- u + (1 - u^{2}) \cdot 2

= - u + 2 (1 - u^{2})

Expand

2 (1 - u^{2}) : 2 - 2 u^{2}

2 (1 - u^{2})

Apply the distributive law:

a (b - c) = ab - a c

a = 2, b = 1, c = u^{2}

= 2 \cdot 1 - 2 u^{2}

Multiply the numbers:

2 \cdot 1 = 2

= 2 - 2 u^{2}

= - u + 2 - 2 u^{2}

- u + 2 - 2 u^{2} = 0

Write in the standard form

a x^{2} + b x + c = 0

- 2 u^{2} - u + 2 = 0

Solve with the quadratic formula

- 2 u^{2} - u + 2 = 0

Quadratic Equation Formula:

For

a = - 2, b = - 1, c = 2

u_{1, 2} = \frac{- ( - 1 ) \pm ( - 1 ) ^{2} - 4 ( - 2 ) \cdot 2}{2 ( - 2 )}

u_{1, 2} = \frac{- ( - 1 ) \pm ( - 1 ) ^{2} - 4 ( - 2 ) \cdot 2}{2 ( - 2 )}

(- 1)^{2} - 4 (- 2) \cdot 2 = 17

(- 1)^{2} - 4 (- 2) \cdot 2

Apply rule

- (- a) = a

= (- 1)^{2} + 4 \cdot 2 \cdot 2

(- 1)^{2} = 1

(- 1)^{2}

Apply exponent rule:

(- a)^{n} = a^{n},

n

is even

(- 1)^{2} = 1^{2}

= 1^{2}

Apply rule

1^{a} = 1

= 1

4 \cdot 2 \cdot 2 = 16

4 \cdot 2 \cdot 2

Multiply the numbers:

4 \cdot 2 \cdot 2 = 16

= 16

= 1 + 16

Add the numbers:

1 + 16 = 17

= 17

u_{1, 2} = \frac{- ( - 1 ) \pm 17}{2 ( - 2 )}

Separate the solutions

u_{1} = \frac{- ( - 1 ) + 17}{2 ( - 2 )}, u_{2} = \frac{- ( - 1 ) - 17}{2 ( - 2 )}

u = \frac{- ( - 1 ) + 17}{2 ( - 2 )} : - \frac{1 + 17}{4}

\frac{- ( - 1 ) + 17}{2 ( - 2 )}

Remove parentheses:

(- a) = - a, - (- a) = a

= \frac{1 + 17}{- 2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{1 + 17}{- 4}

Apply the fraction rule:

\frac{a}{- b} = - \frac{a}{b}

= - \frac{1 + 17}{4}

u = \frac{- ( - 1 ) - 17}{2 ( - 2 )} : \frac{17 - 1}{4}

\frac{- ( - 1 ) - 17}{2 ( - 2 )}

Remove parentheses:

(- a) = - a, - (- a) = a

= \frac{1 - 17}{- 2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{1 - 17}{- 4}

Apply the fraction rule:

\frac{- a}{- b} = \frac{a}{b}

1 - 17 = - (17 - 1)

= \frac{17 - 1}{4}

The solutions to the quadratic equation are:

u = - \frac{1 + 17}{4}, u = \frac{17 - 1}{4}

Substitute back

u = sin (θ)

sin (θ) = - \frac{1 + 17}{4}, sin (θ) = \frac{17 - 1}{4}

sin (θ) = - \frac{1 + 17}{4}, sin (θ) = \frac{17 - 1}{4}

sin (θ) = - \frac{1 + 17}{4} :

No Solution

sin (θ) = - \frac{1 + 17}{4}

- 1 \leq sin (x) \leq 1

No Solution

sin (θ) = \frac{17 - 1}{4} : θ = arcsin (\frac{17 - 1}{4}) + 2 πn, θ = π - arcsin (\frac{17 - 1}{4}) + 2 πn

sin (θ) = \frac{17 - 1}{4}

Apply trig inverse properties

sin (θ) = \frac{17 - 1}{4}

General solutions for

sin (θ) = \frac{17 - 1}{4}

sin (x) = a \Rightarrow x = arcsin (a) + 2 πn, x = π - arcsin (a) + 2 πn

θ = arcsin (\frac{17 - 1}{4}) + 2 πn, θ = π - arcsin (\frac{17 - 1}{4}) + 2 πn

θ = arcsin (\frac{17 - 1}{4}) + 2 πn, θ = π - arcsin (\frac{17 - 1}{4}) + 2 πn

Combine all the solutions

θ = arcsin (\frac{17 - 1}{4}) + 2 πn, θ = π - arcsin (\frac{17 - 1}{4}) + 2 πn

Combine all the solutions

θ = 2 πn, θ = π + 2 πn, θ = arcsin (\frac{17 - 1}{4}) + 2 πn, θ = π - arcsin (\frac{17 - 1}{4}) + 2 πn

Show solutions in decimal form

θ = 2 πn, θ = π + 2 πn, θ = 0.89590 \dots + 2 πn, θ = π - 0.89590 \dots + 2 πn

Popular Examples

sin(θ)-sqrt(3)cos(θ)=1

sin (θ) - 3 cos (θ) = 1

cos(a)= 1/2

cos (a) = \frac{1}{2}

cos^2(θ)= 3/4

cos^{2} (θ) = \frac{3}{4}

tan(θ)= 3/8

tan (θ) = \frac{3}{8}

9cos(x)+5=-cos(x)

9 cos (x) + 5 = - cos (x)

Frequently Asked Questions (FAQ)

What is the general solution for 2sin(θ)-tan^2(θ)=0 ?
The general solution for 2sin(θ)-tan^2(θ)=0 is θ=2pin,θ=pi+2pin,θ=0.89590…+2pin,θ=pi-0.89590…+2pin