What is the general solution for 1/(cos(2x))+tan(2x)=3cos(2x),0<x<90 ?

Q: What is the general solution for 1/(cos(2x))+tan(2x)=3cos(2x),0<x<90 ?

The general solution for 1/(cos(2x))+tan(2x)=3cos(2x),0<x<90 is x=(0.72972…)/2 ,x=(180-0.72972…)/2

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Popular Trigonometry >

1/(cos(2x))+tan(2x)=3cos(2x),0<x<90

Solution

\frac{1}{cos ( 2 x )} + tan (2 x) = 3 cos (2 x), 0^{\circ} < x < 9 0^{\circ}

Solution

x = \frac{0.72972 \dots}{2}, x = \frac{18 0 ^{\circ} - 0.72972 \dots}{2}

Radians

x = \frac{0.72972 \dots}{2}, x = \frac{π - 0.72972 \dots}{2}

Solution steps

\frac{1}{cos ( 2 x )} + tan (2 x) = 3 cos (2 x), 0^{\circ} < x < 9 0^{\circ}

Subtract

3 cos (2 x)

from both sides

\frac{1}{cos ( 2 x )} + tan (2 x) - 3 cos (2 x) = 0

Simplify

\frac{1}{cos ( 2 x )} + tan (2 x) - 3 cos (2 x) : \frac{1 + tan ( 2 x ) cos ( 2 x ) - 3 cos ^{2} ( 2 x )}{cos ( 2 x )}

\frac{1}{cos ( 2 x )} + tan (2 x) - 3 cos (2 x)

Convert element to fraction:

tan (2 x) = \frac{tan ( 2 x ) cos ( 2 x )}{cos ( 2 x )}, 3 cos (2 x) = \frac{3 cos ( 2 x ) cos ( 2 x )}{cos ( 2 x )}

= \frac{1}{cos ( 2 x )} + \frac{tan ( 2 x ) cos ( 2 x )}{cos ( 2 x )} - \frac{3 cos ( 2 x ) cos ( 2 x )}{cos ( 2 x )}

Since the denominators are equal, combine the fractions:

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

= \frac{1 + tan ( 2 x ) cos ( 2 x ) - 3 cos ( 2 x ) cos ( 2 x )}{cos ( 2 x )}

1 + tan (2 x) cos (2 x) - 3 cos (2 x) cos (2 x) = 1 + tan (2 x) cos (2 x) - 3 cos^{2} (2 x)

1 + tan (2 x) cos (2 x) - 3 cos (2 x) cos (2 x)

3 cos (2 x) cos (2 x) = 3 cos^{2} (2 x)

3 cos (2 x) cos (2 x)

Apply exponent rule:

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

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

= 3 cos^{1 + 1} (2 x)

Add the numbers:

1 + 1 = 2

= 3 cos^{2} (2 x)

= 1 + tan (2 x) cos (2 x) - 3 cos^{2} (2 x)

= \frac{1 + tan ( 2 x ) cos ( 2 x ) - 3 cos ^{2} ( 2 x )}{cos ( 2 x )}

\frac{1 + tan ( 2 x ) cos ( 2 x ) - 3 cos ^{2} ( 2 x )}{cos ( 2 x )} = 0

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

1 + tan (2 x) cos (2 x) - 3 cos^{2} (2 x) = 0

Express with sin, cos

1 + \frac{sin ( 2 x )}{cos ( 2 x )} cos (2 x) - 3 cos^{2} (2 x) = 0

Simplify

1 + \frac{sin ( 2 x )}{cos ( 2 x )} cos (2 x) - 3 cos^{2} (2 x) : 1 + sin (2 x) - 3 cos^{2} (2 x)

1 + \frac{sin ( 2 x )}{cos ( 2 x )} cos (2 x) - 3 cos^{2} (2 x)

\frac{sin ( 2 x )}{cos ( 2 x )} cos (2 x) = sin (2 x)

\frac{sin ( 2 x )}{cos ( 2 x )} cos (2 x)

Multiply fractions:

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

= \frac{sin ( 2 x ) cos ( 2 x )}{cos ( 2 x )}

Cancel the common factor:

cos (2 x)

= sin (2 x)

= 1 + sin (2 x) - 3 cos^{2} (2 x)

1 + sin (2 x) - 3 cos^{2} (2 x) = 0

Add

3 cos^{2} (2 x)

to both sides

1 + sin (2 x) = 3 cos^{2} (2 x)

Square both sides

(1 + sin (2 x))^{2} = (3 cos^{2} (2 x))^{2}

Subtract

(3 cos^{2} (2 x))^{2}

from both sides

(1 + sin (2 x))^{2} - 9 cos^{4} (2 x) = 0

Factor

(1 + sin (2 x))^{2} - 9 cos^{4} (2 x) : (1 + sin (2 x) + 3 cos^{2} (2 x)) (1 + sin (2 x) - 3 cos^{2} (2 x))

(1 + sin (2 x))^{2} - 9 cos^{4} (2 x)

Rewrite

(1 + sin (2 x))^{2} - 9 cos^{4} (2 x)

(1 + sin (2 x))^{2} - (3 cos^{2} (2 x))^{2}

(1 + sin (2 x))^{2} - 9 cos^{4} (2 x)

Rewrite

9

3^{2}

= (1 + sin (2 x))^{2} - 3^{2} cos^{4} (2 x)

Apply exponent rule:

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

cos^{4} (2 x) = (cos^{2} (2 x))^{2}

= (1 + sin (2 x))^{2} - 3^{2} (cos^{2} (2 x))^{2}

Apply exponent rule:

a^{m} b^{m} = (ab)^{m}

3^{2} (cos^{2} (2 x))^{2} = (3 cos^{2} (2 x))^{2}

= (1 + sin (2 x))^{2} - (3 cos^{2} (2 x))^{2}

= (1 + sin (2 x))^{2} - (3 cos^{2} (2 x))^{2}

Apply Difference of Two Squares Formula:

x^{2} - y^{2} = (x + y) (x - y)

(1 + sin (2 x))^{2} - (3 cos^{2} (2 x))^{2} = ((1 + sin (2 x)) + 3 cos^{2} (2 x)) ((1 + sin (2 x)) - 3 cos^{2} (2 x))

= ((1 + sin (2 x)) + 3 cos^{2} (2 x)) ((1 + sin (2 x)) - 3 cos^{2} (2 x))

Refine

= (3 cos^{2} (2 x) + sin (2 x) + 1) (sin (2 x) - 3 cos^{2} (2 x) + 1)

(1 + sin (2 x) + 3 cos^{2} (2 x)) (1 + sin (2 x) - 3 cos^{2} (2 x)) = 0

Solving each part separately

1 + sin (2 x) + 3 cos^{2} (2 x) = 0 or 1 + sin (2 x) - 3 cos^{2} (2 x) = 0

1 + sin (2 x) + 3 cos^{2} (2 x) = 0, 0 < x < 9 0^{\circ} :

No Solution

1 + sin (2 x) + 3 cos^{2} (2 x) = 0, 0 < x < 9 0^{\circ}

Rewrite using trig identities

1 + sin (2 x) + 3 cos^{2} (2 x)

Use the Pythagorean identity:

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

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

= 1 + sin (2 x) + 3 (1 - sin^{2} (2 x))

Simplify

1 + sin (2 x) + 3 (1 - sin^{2} (2 x)) : sin (2 x) - 3 sin^{2} (2 x) + 4

1 + sin (2 x) + 3 (1 - sin^{2} (2 x))

Expand

3 (1 - sin^{2} (2 x)) : 3 - 3 sin^{2} (2 x)

3 (1 - sin^{2} (2 x))

Apply the distributive law:

a (b - c) = ab - a c

a = 3, b = 1, c = sin^{2} (2 x)

= 3 \cdot 1 - 3 sin^{2} (2 x)

Multiply the numbers:

3 \cdot 1 = 3

= 3 - 3 sin^{2} (2 x)

= 1 + sin (2 x) + 3 - 3 sin^{2} (2 x)

Simplify

1 + sin (2 x) + 3 - 3 sin^{2} (2 x) : sin (2 x) - 3 sin^{2} (2 x) + 4

1 + sin (2 x) + 3 - 3 sin^{2} (2 x)

Group like terms

= sin (2 x) - 3 sin^{2} (2 x) + 1 + 3

Add the numbers:

1 + 3 = 4

= sin (2 x) - 3 sin^{2} (2 x) + 4

= sin (2 x) - 3 sin^{2} (2 x) + 4

= sin (2 x) - 3 sin^{2} (2 x) + 4

4 + sin (2 x) - 3 sin^{2} (2 x) = 0

Solve by substitution

4 + sin (2 x) - 3 sin^{2} (2 x) = 0

Let:

sin (2 x) = u

4 + u - 3 u^{2} = 0

4 + u - 3 u^{2} = 0 : u = - 1, u = \frac{4}{3}

4 + u - 3 u^{2} = 0

Write in the standard form

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

- 3 u^{2} + u + 4 = 0

Solve with the quadratic formula

- 3 u^{2} + u + 4 = 0

Quadratic Equation Formula:

For

a = - 3, b = 1, c = 4

u_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 ( - 3 ) \cdot 4}{2 ( - 3 )}

u_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 ( - 3 ) \cdot 4}{2 ( - 3 )}

1^{2} - 4 (- 3) \cdot 4 = 7

1^{2} - 4 (- 3) \cdot 4

Apply rule

1^{a} = 1

1^{2} = 1

= 1 - 4 (- 3) \cdot 4

Apply rule

- (- a) = a

= 1 + 4 \cdot 3 \cdot 4

Multiply the numbers:

4 \cdot 3 \cdot 4 = 48

= 1 + 48

Add the numbers:

1 + 48 = 49

= 49

Factor the number:

49 = 7^{2}

= 7^{2}

Apply radical rule:

7^{2} = 7

= 7

u_{1, 2} = \frac{- 1 \pm 7}{2 ( - 3 )}

Separate the solutions

u_{1} = \frac{- 1 + 7}{2 ( - 3 )}, u_{2} = \frac{- 1 - 7}{2 ( - 3 )}

u = \frac{- 1 + 7}{2 ( - 3 )} : - 1

\frac{- 1 + 7}{2 ( - 3 )}

Remove parentheses:

(- a) = - a

= \frac{- 1 + 7}{- 2 \cdot 3}

Add/Subtract the numbers:

- 1 + 7 = 6

= \frac{6}{- 2 \cdot 3}

Multiply the numbers:

2 \cdot 3 = 6

= \frac{6}{- 6}

Apply the fraction rule:

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

= - \frac{6}{6}

Apply rule

\frac{a}{a} = 1

= - 1

u = \frac{- 1 - 7}{2 ( - 3 )} : \frac{4}{3}

\frac{- 1 - 7}{2 ( - 3 )}

Remove parentheses:

(- a) = - a

= \frac{- 1 - 7}{- 2 \cdot 3}

Subtract the numbers:

- 1 - 7 = - 8

= \frac{- 8}{- 2 \cdot 3}

Multiply the numbers:

2 \cdot 3 = 6

= \frac{- 8}{- 6}

Apply the fraction rule:

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

= \frac{8}{6}

Cancel the common factor:

2

= \frac{4}{3}

The solutions to the quadratic equation are:

u = - 1, u = \frac{4}{3}

Substitute back

u = sin (2 x)

sin (2 x) = - 1, sin (2 x) = \frac{4}{3}

sin (2 x) = - 1, sin (2 x) = \frac{4}{3}

sin (2 x) = - 1, 0 < x < 9 0^{\circ} :

No Solution

sin (2 x) = - 1, 0 < x < 9 0^{\circ}

General solutions for

sin (2 x) = - 1

sin (x)

periodicity table with

36 0^{\circ} n

cycle:

x 0 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ} 12 0^{\circ} 13 5^{\circ} 15 0^{\circ} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x 18 0^{\circ} 21 0^{\circ} 22 5^{\circ} 24 0^{\circ} 27 0^{\circ} 30 0^{\circ} 31 5^{\circ} 33 0^{\circ} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

2 x = 27 0^{\circ} + 36 0^{\circ} n

2 x = 27 0^{\circ} + 36 0^{\circ} n

Solve

2 x = 27 0^{\circ} + 36 0^{\circ} n : x = 13 5^{\circ} + 18 0^{\circ} n

2 x = 27 0^{\circ} + 36 0^{\circ} n

Divide both sides by

2

2 x = 27 0^{\circ} + 36 0^{\circ} n

Divide both sides by

2

\frac{2 x}{2} = \frac{27 0 ^{\circ}}{2} + \frac{36 0 ^{\circ} n}{2}

Simplify

\frac{2 x}{2} = \frac{27 0 ^{\circ}}{2} + \frac{36 0 ^{\circ} n}{2}

Simplify

\frac{2 x}{2} : x

\frac{2 x}{2}

Divide the numbers:

\frac{2}{2} = 1

= x

Simplify

\frac{27 0 ^{\circ}}{2} + \frac{36 0 ^{\circ} n}{2} : 13 5^{\circ} + 18 0^{\circ} n

\frac{27 0 ^{\circ}}{2} + \frac{36 0 ^{\circ} n}{2}

\frac{27 0 ^{\circ}}{2} = 13 5^{\circ}

\frac{27 0 ^{\circ}}{2}

Apply the fraction rule:

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

= \frac{54 0 ^{\circ}}{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= 13 5^{\circ}

\frac{36 0 ^{\circ} n}{2} = 18 0^{\circ} n

\frac{36 0 ^{\circ} n}{2}

Divide the numbers:

\frac{2}{2} = 1

= 18 0^{\circ} n

= 13 5^{\circ} + 18 0^{\circ} n

x = 13 5^{\circ} + 18 0^{\circ} n

x = 13 5^{\circ} + 18 0^{\circ} n

x = 13 5^{\circ} + 18 0^{\circ} n

x = 13 5^{\circ} + 18 0^{\circ} n

Solutions for the range

0 < x < 9 0^{\circ}

No Solution

sin (2 x) = \frac{4}{3}, 0 < x < 9 0^{\circ} :

No Solution

sin (2 x) = \frac{4}{3}, 0 < x < 9 0^{\circ}

- 1 \leq sin (x) \leq 1

No Solution

Combine all the solutions

No Solution

1 + sin (2 x) - 3 cos^{2} (2 x) = 0, 0 < x < 9 0^{\circ} : x = \frac{arcsin ( \frac{2}{3} )}{2}, x = \frac{18 0 ^{\circ} - arcsin ( \frac{2}{3} )}{2}

1 + sin (2 x) - 3 cos^{2} (2 x) = 0, 0 < x < 9 0^{\circ}

Rewrite using trig identities

1 + sin (2 x) - 3 cos^{2} (2 x)

Use the Pythagorean identity:

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

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

= 1 + sin (2 x) - 3 (1 - sin^{2} (2 x))

Simplify

1 + sin (2 x) - 3 (1 - sin^{2} (2 x)) : 3 sin^{2} (2 x) + sin (2 x) - 2

1 + sin (2 x) - 3 (1 - sin^{2} (2 x))

Expand

- 3 (1 - sin^{2} (2 x)) : - 3 + 3 sin^{2} (2 x)

- 3 (1 - sin^{2} (2 x))

Apply the distributive law:

a (b - c) = ab - a c

a = - 3, b = 1, c = sin^{2} (2 x)

= - 3 \cdot 1 - (- 3) sin^{2} (2 x)

Apply minus-plus rules

- (- a) = a

= - 3 \cdot 1 + 3 sin^{2} (2 x)

Multiply the numbers:

3 \cdot 1 = 3

= - 3 + 3 sin^{2} (2 x)

= 1 + sin (2 x) - 3 + 3 sin^{2} (2 x)

Simplify

1 + sin (2 x) - 3 + 3 sin^{2} (2 x) : 3 sin^{2} (2 x) + sin (2 x) - 2

1 + sin (2 x) - 3 + 3 sin^{2} (2 x)

Group like terms

= sin (2 x) + 3 sin^{2} (2 x) + 1 - 3

Add/Subtract the numbers:

1 - 3 = - 2

= 3 sin^{2} (2 x) + sin (2 x) - 2

= 3 sin^{2} (2 x) + sin (2 x) - 2

= 3 sin^{2} (2 x) + sin (2 x) - 2

- 2 + sin (2 x) + 3 sin^{2} (2 x) = 0

Solve by substitution

- 2 + sin (2 x) + 3 sin^{2} (2 x) = 0

Let:

sin (2 x) = u

- 2 + u + 3 u^{2} = 0

- 2 + u + 3 u^{2} = 0 : u = \frac{2}{3}, u = - 1

- 2 + u + 3 u^{2} = 0

Write in the standard form

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

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

a = 3, b = 1, c = - 2

u_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot 3 ( - 2 )}{2 \cdot 3}

u_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot 3 ( - 2 )}{2 \cdot 3}

1^{2} - 4 \cdot 3 (- 2) = 5

1^{2} - 4 \cdot 3 (- 2)

Apply rule

1^{a} = 1

1^{2} = 1

= 1 - 4 \cdot 3 (- 2)

Apply rule

- (- a) = a

= 1 + 4 \cdot 3 \cdot 2

Multiply the numbers:

4 \cdot 3 \cdot 2 = 24

= 1 + 24

Add the numbers:

1 + 24 = 25

= 25

Factor the number:

25 = 5^{2}

= 5^{2}

Apply radical rule:

5^{2} = 5

= 5

u_{1, 2} = \frac{- 1 \pm 5}{2 \cdot 3}

Separate the solutions

u_{1} = \frac{- 1 + 5}{2 \cdot 3}, u_{2} = \frac{- 1 - 5}{2 \cdot 3}

u = \frac{- 1 + 5}{2 \cdot 3} : \frac{2}{3}

\frac{- 1 + 5}{2 \cdot 3}

Add/Subtract the numbers:

- 1 + 5 = 4

= \frac{4}{2 \cdot 3}

Multiply the numbers:

2 \cdot 3 = 6

= \frac{4}{6}

Cancel the common factor:

2

= \frac{2}{3}

u = \frac{- 1 - 5}{2 \cdot 3} : - 1

\frac{- 1 - 5}{2 \cdot 3}

Subtract the numbers:

- 1 - 5 = - 6

= \frac{- 6}{2 \cdot 3}

Multiply the numbers:

2 \cdot 3 = 6

= \frac{- 6}{6}

Apply the fraction rule:

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

= - \frac{6}{6}

Apply rule

\frac{a}{a} = 1

= - 1

The solutions to the quadratic equation are:

u = \frac{2}{3}, u = - 1

Substitute back

u = sin (2 x)

sin (2 x) = \frac{2}{3}, sin (2 x) = - 1

sin (2 x) = \frac{2}{3}, sin (2 x) = - 1

sin (2 x) = \frac{2}{3}, 0 < x < 9 0^{\circ} : x = \frac{arcsin ( \frac{2}{3} )}{2}, x = \frac{18 0 ^{\circ} - arcsin ( \frac{2}{3} )}{2}

sin (2 x) = \frac{2}{3}, 0 < x < 9 0^{\circ}

Apply trig inverse properties

sin (2 x) = \frac{2}{3}

General solutions for

sin (2 x) = \frac{2}{3}

sin (x) = a \Rightarrow x = arcsin (a) + 36 0^{\circ} n, x = 18 0^{\circ} - arcsin (a) + 36 0^{\circ} n

2 x = arcsin (\frac{2}{3}) + 36 0^{\circ} n, 2 x = 18 0^{\circ} - arcsin (\frac{2}{3}) + 36 0^{\circ} n

2 x = arcsin (\frac{2}{3}) + 36 0^{\circ} n, 2 x = 18 0^{\circ} - arcsin (\frac{2}{3}) + 36 0^{\circ} n

Solve

2 x = arcsin (\frac{2}{3}) + 36 0^{\circ} n : x = \frac{arcsin ( \frac{2}{3} )}{2} + 18 0^{\circ} n

2 x = arcsin (\frac{2}{3}) + 36 0^{\circ} n

Divide both sides by

2

2 x = arcsin (\frac{2}{3}) + 36 0^{\circ} n

Divide both sides by

2

\frac{2 x}{2} = \frac{arcsin ( \frac{2}{3} )}{2} + \frac{36 0 ^{\circ} n}{2}

Simplify

x = \frac{arcsin ( \frac{2}{3} )}{2} + 18 0^{\circ} n

x = \frac{arcsin ( \frac{2}{3} )}{2} + 18 0^{\circ} n

Solve

2 x = 18 0^{\circ} - arcsin (\frac{2}{3}) + 36 0^{\circ} n : x = 9 0^{\circ} - \frac{arcsin ( \frac{2}{3} )}{2} + 18 0^{\circ} n

2 x = 18 0^{\circ} - arcsin (\frac{2}{3}) + 36 0^{\circ} n

Divide both sides by

2

2 x = 18 0^{\circ} - arcsin (\frac{2}{3}) + 36 0^{\circ} n

Divide both sides by

2

\frac{2 x}{2} = 9 0^{\circ} - \frac{arcsin ( \frac{2}{3} )}{2} + \frac{36 0 ^{\circ} n}{2}

Simplify

x = 9 0^{\circ} - \frac{arcsin ( \frac{2}{3} )}{2} + 18 0^{\circ} n

x = 9 0^{\circ} - \frac{arcsin ( \frac{2}{3} )}{2} + 18 0^{\circ} n

x = \frac{arcsin ( \frac{2}{3} )}{2} + 18 0^{\circ} n, x = 9 0^{\circ} - \frac{arcsin ( \frac{2}{3} )}{2} + 18 0^{\circ} n

Solutions for the range

0 < x < 9 0^{\circ}

x = \frac{arcsin ( \frac{2}{3} )}{2}, x = \frac{18 0 ^{\circ} - arcsin ( \frac{2}{3} )}{2}

sin (2 x) = - 1, 0 < x < 9 0^{\circ} :

No Solution

sin (2 x) = - 1, 0 < x < 9 0^{\circ}

General solutions for

sin (2 x) = - 1

sin (x)

periodicity table with

36 0^{\circ} n

cycle:

x 0 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ} 12 0^{\circ} 13 5^{\circ} 15 0^{\circ} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x 18 0^{\circ} 21 0^{\circ} 22 5^{\circ} 24 0^{\circ} 27 0^{\circ} 30 0^{\circ} 31 5^{\circ} 33 0^{\circ} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

2 x = 27 0^{\circ} + 36 0^{\circ} n

2 x = 27 0^{\circ} + 36 0^{\circ} n

Solve

2 x = 27 0^{\circ} + 36 0^{\circ} n : x = 13 5^{\circ} + 18 0^{\circ} n

2 x = 27 0^{\circ} + 36 0^{\circ} n

Divide both sides by

2

2 x = 27 0^{\circ} + 36 0^{\circ} n

Divide both sides by

2

\frac{2 x}{2} = \frac{27 0 ^{\circ}}{2} + \frac{36 0 ^{\circ} n}{2}

Simplify

\frac{2 x}{2} = \frac{27 0 ^{\circ}}{2} + \frac{36 0 ^{\circ} n}{2}

Simplify

\frac{2 x}{2} : x

\frac{2 x}{2}

Divide the numbers:

\frac{2}{2} = 1

= x

Simplify

\frac{27 0 ^{\circ}}{2} + \frac{36 0 ^{\circ} n}{2} : 13 5^{\circ} + 18 0^{\circ} n

\frac{27 0 ^{\circ}}{2} + \frac{36 0 ^{\circ} n}{2}

\frac{27 0 ^{\circ}}{2} = 13 5^{\circ}

\frac{27 0 ^{\circ}}{2}

Apply the fraction rule:

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

= \frac{54 0 ^{\circ}}{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= 13 5^{\circ}

\frac{36 0 ^{\circ} n}{2} = 18 0^{\circ} n

\frac{36 0 ^{\circ} n}{2}

Divide the numbers:

\frac{2}{2} = 1

= 18 0^{\circ} n

= 13 5^{\circ} + 18 0^{\circ} n

x = 13 5^{\circ} + 18 0^{\circ} n

x = 13 5^{\circ} + 18 0^{\circ} n

x = 13 5^{\circ} + 18 0^{\circ} n

x = 13 5^{\circ} + 18 0^{\circ} n

Solutions for the range

0 < x < 9 0^{\circ}

No Solution

Combine all the solutions

x = \frac{arcsin ( \frac{2}{3} )}{2}, x = \frac{18 0 ^{\circ} - arcsin ( \frac{2}{3} )}{2}

Combine all the solutions

x = \frac{arcsin ( \frac{2}{3} )}{2}, x = \frac{18 0 ^{\circ} - arcsin ( \frac{2}{3} )}{2}

Verify solutions by plugging them into the original equation

Check the solutions by plugging them into

\frac{1}{cos ( 2 x )} + tan (2 x) = 3 cos (2 x)

Remove the ones that don't agree with the equation.

Check the solution

\frac{arcsin ( \frac{2}{3} )}{2} :

True

\frac{arcsin ( \frac{2}{3} )}{2}

Plug in

n = 1

\frac{arcsin ( \frac{2}{3} )}{2}

For

\frac{1}{cos ( 2 x )} + tan (2 x) = 3 cos (2 x)

plug in

x = \frac{arcsin ( \frac{2}{3} )}{2}

\frac{1}{cos ( 2 \cdot \frac{a r c s i n ( \frac{2}{3} )}{2} )} + tan (2 \cdot \frac{arcsin ( \frac{2}{3} )}{2}) = 3 cos (2 \cdot \frac{arcsin ( \frac{2}{3} )}{2})

Refine

2.23606 \dots = 2.23606 \dots

\Rightarrow True

Check the solution

\frac{18 0 ^{\circ} - arcsin ( \frac{2}{3} )}{2} :

True

\frac{18 0 ^{\circ} - arcsin ( \frac{2}{3} )}{2}

Plug in

n = 1

\frac{18 0 ^{\circ} - arcsin ( \frac{2}{3} )}{2}

For

\frac{1}{cos ( 2 x )} + tan (2 x) = 3 cos (2 x)

plug in

x = \frac{18 0 ^{\circ} - arcsin ( \frac{2}{3} )}{2}

\frac{1}{cos ( 2 \cdot \frac{18 0 ^{\circ} - a r c s i n ( \frac{2}{3} )}{2} )} + tan (2 \cdot \frac{18 0 ^{\circ} - arcsin ( \frac{2}{3} )}{2}) = 3 cos (2 \cdot \frac{18 0 ^{\circ} - arcsin ( \frac{2}{3} )}{2})

Refine

- 2.23606 \dots = - 2.23606 \dots

\Rightarrow True

x = \frac{arcsin ( \frac{2}{3} )}{2}, x = \frac{18 0 ^{\circ} - arcsin ( \frac{2}{3} )}{2}

Show solutions in decimal form

x = \frac{0.72972 \dots}{2}, x = \frac{18 0 ^{\circ} - 0.72972 \dots}{2}

Graph

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Frequently Asked Questions (FAQ)

What is the general solution for 1/(cos(2x))+tan(2x)=3cos(2x),0<x<90 ?
The general solution for 1/(cos(2x))+tan(2x)=3cos(2x),0<x<90 is x=(0.72972…)/2 ,x=(180-0.72972…)/2