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Popular >

4sin^2(x)+3tan(x)>sec^2(x)

Solution

4 sin^{2} (x) + 3 tan (x) > sec^{2} (x)

Solution

\frac{π}{12} + πn < x < \frac{5 π}{12} + πn

Interval Notation

(\frac{π}{12} + πn, \frac{5 π}{12} + πn)

Decimal Notation

0.26179 \dots + πn < x < 1.30899 \dots + πn

Solution steps

4 sin^{2} (x) + 3 tan (x) > sec^{2} (x)

Move

sec^{2} (x)

to the left side

4 sin^{2} (x) + 3 tan (x) > sec^{2} (x)

Subtract

sec^{2} (x)

from both sides

4 sin^{2} (x) + 3 tan (x) - sec^{2} (x) > sec^{2} (x) - sec^{2} (x)

4 sin^{2} (x) + 3 tan (x) - sec^{2} (x) > 0

4 sin^{2} (x) + 3 tan (x) - sec^{2} (x) > 0

Use the following identity:

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

Therefore

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

4 (1 - cos^{2} (x)) + 3 tan (x) - sec^{2} (x) > 0

Periodicity of

4 (1 - cos^{2} (x)) + 3 tan (x) - sec^{2} (x) : π

The compound periodicity of the sum of periodic functions is the least common multiplier of the periods

4 (1 - cos^{2} (x)), 3 tan (x), sec^{2} (x)

Periodicity of

4 (1 - cos^{2} (x)) : π

Periodicity of

cos^{n} (x) = \frac{Periodicity of cos ( x )}{2},

if n is even

Periodicity of

cos (x) : 2 π

Periodicity of

cos (x)

2 π

= 2 π

\frac{2 π}{2}

Simplify

π

Periodicity of

3 tan (x) : π

Periodicity of

a \cdot tan (b x + c) + d = \frac{periodicity of tan ( x )}{∣ b ∣}

Periodicity of

tan (x)

π

= \frac{π}{∣ 1 ∣}

Simplify

= π

Periodicity of

sec^{2} (x) : π

Periodicity of

sec^{n} (x) = \frac{Periodicity of sec ( x )}{2},

if n is even

Periodicity of

sec (x) : 2 π

Periodicity of

sec (x)

2 π

= 2 π

\frac{2 π}{2}

Simplify

π

Combine periods:

π, π, π

= π

Express with sin, cos

4 (1 - cos^{2} (x)) + 3 tan (x) - sec^{2} (x) > 0

Use the basic trigonometric identity:

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

4 (1 - cos^{2} (x)) + 3 \cdot \frac{sin ( x )}{cos ( x )} - sec^{2} (x) > 0

Use the basic trigonometric identity:

sec (x) = \frac{1}{cos ( x )}

4 (1 - cos^{2} (x)) + 3 \cdot \frac{sin ( x )}{cos ( x )} - (\frac{1}{cos ( x )})^{2} > 0

4 (1 - cos^{2} (x)) + 3 \cdot \frac{sin ( x )}{cos ( x )} - (\frac{1}{cos ( x )})^{2} > 0

Simplify

4 (1 - cos^{2} (x)) + 3 \cdot \frac{sin ( x )}{cos ( x )} - (\frac{1}{cos ( x )})^{2} : \frac{4 cos ^{2} ( x ) ( 1 - cos ^{2} ( x ) ) + 3 sin ( x ) cos ( x ) - 1}{cos ^{2} ( x )}

4 (1 - cos^{2} (x)) + 3 \cdot \frac{sin ( x )}{cos ( x )} - (\frac{1}{cos ( x )})^{2}

3 \cdot \frac{sin ( x )}{cos ( x )} = \frac{3 sin ( x )}{cos ( x )}

3 \cdot \frac{sin ( x )}{cos ( x )}

Multiply fractions:

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

= \frac{sin ( x ) \cdot 3}{cos ( x )}

(\frac{1}{cos ( x )})^{2} = \frac{1}{cos ^{2} ( x )}

(\frac{1}{cos ( x )})^{2}

Apply exponent rule:

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

= \frac{1 ^{2}}{cos ^{2} ( x )}

Apply rule

1^{a} = 1

1^{2} = 1

= \frac{1}{cos ^{2} ( x )}

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

Convert element to fraction:

4 (- cos^{2} (x) + 1) = \frac{4 ( - cos ^{2} ( x ) + 1 )}{1}

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

Least Common Multiplier of

1, cos (x), cos^{2} (x) : cos^{2} (x)

1, cos (x), cos^{2} (x)

Lowest Common Multiplier (LCM)

Compute an expression comprised of factors that appear in at least one of the factored expressions

= cos^{2} (x)

Adjust Fractions based on the LCM

Multiply each numerator by the same amount needed to multiply its

corresponding denominator to turn it into the LCM

cos^{2} (x)

For

\frac{4 ( 1 - cos ^{2} ( x ) )}{1} :

multiply the denominator and numerator by

cos^{2} (x)

\frac{4 ( 1 - cos ^{2} ( x ) )}{1} = \frac{4 ( 1 - cos ^{2} ( x ) ) cos ^{2} ( x )}{1 \cdot cos ^{2} ( x )} = \frac{4 ( 1 - cos ^{2} ( x ) ) cos ^{2} ( x )}{cos ^{2} ( x )}

For

\frac{sin ( x ) \cdot 3}{cos ( x )} :

multiply the denominator and numerator by

cos (x)

\frac{sin ( x ) \cdot 3}{cos ( x )} = \frac{sin ( x ) \cdot 3 cos ( x )}{cos ( x ) cos ( x )} = \frac{sin ( x ) \cdot 3 cos ( x )}{cos ^{2} ( x )}

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

Since the denominators are equal, combine the fractions:

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

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

\frac{4 cos ^{2} ( x ) ( 1 - cos ^{2} ( x ) ) + 3 sin ( x ) cos ( x ) - 1}{cos ^{2} ( x )} > 0

Find the zeroes and undifined points of

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

for

0 \leq x < π

To find the zeroes, set the inequality to zero

\frac{4 cos ^{2} ( x ) ( 1 - cos ^{2} ( x ) ) + 3 sin ( x ) cos ( x ) - 1}{cos ^{2} ( x )} = 0

\frac{4 cos ^{2} ( x ) ( 1 - cos ^{2} ( x ) ) + 3 sin ( x ) cos ( x ) - 1}{cos ^{2} ( x )} = 0, 0 \leq x < π : x = \frac{π}{12}, x = \frac{5 π}{12}

\frac{4 cos ^{2} ( x ) ( 1 - cos ^{2} ( x ) ) + 3 sin ( x ) cos ( x ) - 1}{cos ^{2} ( x )} = 0, 0 \leq x < π

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

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

- 1 + 3 cos (x) sin (x) + 4 cos^{2} (x) sin^{2} (x) = 0

Factor

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

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

Break the expression into groups

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

Definition

Factors of

4 : 1, 2, 4

4

Divisors (Factors)

Find the Prime factors of

4 : 2, 2

4

4

divides by

24 = 2 \cdot 2

= 2 \cdot 2

2

is a prime number, therefore no further factorization is possible

= 2 \cdot 2

Add the prime factors:

2

Add 1 and the number

4

itself

1, 4

The factors of

4

1, 2, 4

Negative factors of

4 : - 1, - 2, - 4

Multiply the factors by

- 1

to get the negative factors

- 1, - 2, - 4

For every two factors such that

u * v = - 4,

check if

u + v = 3

Check

u = 1, v = - 4 : u * v = - 4, u + v = - 3 \Rightarrow

FalseCheck

u = 2, v = - 2 : u * v = - 4, u + v = 0 \Rightarrow

False

u = 4, v = - 1

Group into

(a x^{2} y^{2} + ux y) + (vx y + c)

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

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

Factor out

sin (x) cos (x)

from

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

4 sin^{2} (x) cos^{2} (x) - sin (x) cos (x)

Apply exponent rule:

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

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

= 4 sin (x) sin (x) cos (x) cos (x) - sin (x) cos (x)

Factor out common term

sin (x) cos (x)

= sin (x) cos (x) (4 sin (x) cos (x) - 1)

= sin (x) cos (x) (4 sin (x) cos (x) - 1) + (4 sin (x) cos (x) - 1)

Factor out common term

4 sin (x) cos (x) - 1

= (4 sin (x) cos (x) - 1) (sin (x) cos (x) + 1)

(4 sin (x) cos (x) - 1) (sin (x) cos (x) + 1) = 0

Solving each part separately

4 sin (x) cos (x) - 1 = 0 or sin (x) cos (x) + 1 = 0

4 sin (x) cos (x) - 1 = 0, 0 \leq x < π : x = \frac{π}{12}, x = \frac{5 π}{12}

4 sin (x) cos (x) - 1 = 0, 0 \leq x < π

Rewrite using trig identities

4 sin (x) cos (x) - 1

Use the Double Angle identity:

2 sin (x) cos (x) = sin (2 x)

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

= - 1 + 4 \cdot \frac{sin ( 2 x )}{2}

- 1 + 4 \cdot \frac{sin ( 2 x )}{2} = 0

4 \cdot \frac{sin ( 2 x )}{2} = 2 sin (2 x)

4 \cdot \frac{sin ( 2 x )}{2}

Multiply fractions:

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

= \frac{sin ( 2 x ) \cdot 4}{2}

Divide the numbers:

\frac{4}{2} = 2

= 2 sin (2 x)

- 1 + 2 sin (2 x) = 0

Move

1

to the right side

- 1 + 2 sin (2 x) = 0

Add

1

to both sides

- 1 + 2 sin (2 x) + 1 = 0 + 1

Simplify

2 sin (2 x) = 1

2 sin (2 x) = 1

Divide both sides by

2

2 sin (2 x) = 1

Divide both sides by

2

\frac{2 sin ( 2 x )}{2} = \frac{1}{2}

Simplify

sin (2 x) = \frac{1}{2}

sin (2 x) = \frac{1}{2}

General solutions for

sin (2 x) = \frac{1}{2}

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}

2 x = \frac{π}{6} + 2 πn, 2 x = \frac{5 π}{6} + 2 πn

2 x = \frac{π}{6} + 2 πn, 2 x = \frac{5 π}{6} + 2 πn

Solve

2 x = \frac{π}{6} + 2 πn : x = \frac{π}{12} + πn

2 x = \frac{π}{6} + 2 πn

Divide both sides by

2

2 x = \frac{π}{6} + 2 πn

Divide both sides by

2

\frac{2 x}{2} = \frac{\frac{π}{6}}{2} + \frac{2 πn}{2}

Simplify

\frac{2 x}{2} = \frac{\frac{π}{6}}{2} + \frac{2 πn}{2}

Simplify

\frac{2 x}{2} : x

\frac{2 x}{2}

Divide the numbers:

\frac{2}{2} = 1

= x

Simplify

\frac{\frac{π}{6}}{2} + \frac{2 πn}{2} : \frac{π}{12} + πn

\frac{\frac{π}{6}}{2} + \frac{2 πn}{2}

\frac{\frac{π}{6}}{2} = \frac{π}{12}

\frac{\frac{π}{6}}{2}

Apply the fraction rule:

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

= \frac{π}{6 \cdot 2}

Multiply the numbers:

6 \cdot 2 = 12

= \frac{π}{12}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

Divide the numbers:

\frac{2}{2} = 1

= πn

= \frac{π}{12} + πn

x = \frac{π}{12} + πn

x = \frac{π}{12} + πn

x = \frac{π}{12} + πn

Solve

2 x = \frac{5 π}{6} + 2 πn : x = \frac{5 π}{12} + πn

2 x = \frac{5 π}{6} + 2 πn

Divide both sides by

2

2 x = \frac{5 π}{6} + 2 πn

Divide both sides by

2

\frac{2 x}{2} = \frac{\frac{5 π}{6}}{2} + \frac{2 πn}{2}

Simplify

\frac{2 x}{2} = \frac{\frac{5 π}{6}}{2} + \frac{2 πn}{2}

Simplify

\frac{2 x}{2} : x

\frac{2 x}{2}

Divide the numbers:

\frac{2}{2} = 1

= x

Simplify

\frac{\frac{5 π}{6}}{2} + \frac{2 πn}{2} : \frac{5 π}{12} + πn

\frac{\frac{5 π}{6}}{2} + \frac{2 πn}{2}

\frac{\frac{5 π}{6}}{2} = \frac{5 π}{12}

\frac{\frac{5 π}{6}}{2}

Apply the fraction rule:

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

= \frac{5 π}{6 \cdot 2}

Multiply the numbers:

6 \cdot 2 = 12

= \frac{5 π}{12}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

Divide the numbers:

\frac{2}{2} = 1

= πn

= \frac{5 π}{12} + πn

x = \frac{5 π}{12} + πn

x = \frac{5 π}{12} + πn

x = \frac{5 π}{12} + πn

x = \frac{π}{12} + πn, x = \frac{5 π}{12} + πn

Solutions for the range

0 \leq x < π

x = \frac{π}{12}, x = \frac{5 π}{12}

sin (x) cos (x) + 1 = 0, 0 \leq x < π :

No Solution

sin (x) cos (x) + 1 = 0, 0 \leq x < π

Rewrite using trig identities

sin (x) cos (x) + 1

Use the Double Angle identity:

2 sin (x) cos (x) = sin (2 x)

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

= 1 + \frac{sin ( 2 x )}{2}

1 + \frac{sin ( 2 x )}{2} = 0

Move

1

to the right side

1 + \frac{sin ( 2 x )}{2} = 0

Subtract

1

from both sides

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

Simplify

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

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

Multiply both sides by

2

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

Multiply both sides by

2

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

Simplify

sin (2 x) = - 2

sin (2 x) = - 2

- 1 \leq sin (x) \leq 1

No Solution

Combine all the solutions

x = \frac{π}{12}, x = \frac{5 π}{12}

Find the undefined points:

x = \frac{π}{2}

Find the zeros of the denominator

cos^{2} (x) = 0

Apply rule

x^{n} = 0 \Rightarrow x = 0

cos (x) = 0

General solutions for

cos (x) = 0

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

x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

Solutions for the range

0 \leq x < π

x = \frac{π}{2}

\frac{π}{12}, \frac{5 π}{12}, \frac{π}{2}

Identify the intervals

0 < x < \frac{π}{12}, \frac{π}{12} < x < \frac{5 π}{12}, \frac{5 π}{12} < x < \frac{π}{2}, \frac{π}{2} < x < π

Summarize in a table:

4 cos^{2} (x) (1 - cos^{2} (x)) + 3 sin (x) cos (x) - 1 cos^{2} (x) \frac{4 c o s ^{2} ( x ) ( 1 - c o s ^{2} ( x ) ) + 3 s i n ( x ) c o s ( x ) - 1}{c o s ^{2} ( x )} x = 0 - + - 0 < x < \frac{π}{12} - + - x = \frac{π}{12} 0 + 0 \frac{π}{12} < x < \frac{5 π}{12} + + + x = \frac{5 π}{12} 0 + 0 \frac{5 π}{12} < x < \frac{π}{2} - + - x = \frac{π}{2} - 0 Undefined \frac{π}{2} < x < π - + - x = π - + -

Identify the intervals that satisfy the required condition:

> 0

\frac{π}{12} < x < \frac{5 π}{12}

Apply the periodicity of

4 (1 - cos^{2} (x)) + 3 tan (x) - sec^{2} (x)

\frac{π}{12} + πn < x < \frac{5 π}{12} + πn

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sin(pi/6)cos(x)+cos(pi/6)sin(x)<= 1

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1/(sqrt(3))<tan(x)

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