English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

1/(cos^2(x))= 4/3

Solution

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

Solution

x = \frac{π}{6} + 2 πn, x = \frac{11 π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn, x = \frac{7 π}{6} + 2 πn

Degrees

x = 3 0^{\circ} + 36 0^{\circ} n, x = 33 0^{\circ} + 36 0^{\circ} n, x = 15 0^{\circ} + 36 0^{\circ} n, x = 21 0^{\circ} + 36 0^{\circ} n

Solution steps

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

Solve by substitution

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

Let:

cos (x) = u

\frac{1}{u ^{2}} = \frac{4}{3}

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

\frac{1}{u ^{2}} = \frac{4}{3}

Cross multiply

\frac{1}{u ^{2}} = \frac{4}{3}

Apply fraction cross multiply: if

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

then

a \cdot d = b \cdot c

1 \cdot 3 = u^{2} \cdot 4

Simplify

1 \cdot 3 : 3

1 \cdot 3

Multiply the numbers:

1 \cdot 3 = 3

= 3

3 = u^{2} \cdot 4

3 = u^{2} \cdot 4

Solve

3 = u^{2} \cdot 4 : u = \frac{3}{2}, u = - \frac{3}{2}

3 = u^{2} \cdot 4

Switch sides

u^{2} \cdot 4 = 3

Divide both sides by

4

u^{2} \cdot 4 = 3

Divide both sides by

4

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

Simplify

u^{2} = \frac{3}{4}

u^{2} = \frac{3}{4}

For

x^{2} = f (a)

the solutions are

x = f (a), - f (a)

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

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

\frac{3}{4}

Apply radical rule:

assuming

a \geq 0, b \geq 0

= \frac{3}{4}

4 = 2

4

Factor the number:

4 = 2^{2}

= 2^{2}

Apply radical rule:

2^{2} = 2

= 2

= \frac{3}{2}

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

- \frac{3}{4}

Simplify

\frac{3}{4} : \frac{3}{2}

\frac{3}{4}

Apply radical rule:

assuming

a \geq 0, b \geq 0

= \frac{3}{4}

4 = 2

4

Factor the number:

4 = 2^{2}

= 2^{2}

Apply radical rule:

2^{2} = 2

= 2

= \frac{3}{2}

= - \frac{3}{2}

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

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

Verify Solutions

Find undefined (singularity) points:

u = 0

Take the denominator(s) of

\frac{1}{u ^{2}}

and compare to zero

Solve

u^{2} = 0 : u = 0

u^{2} = 0

Apply rule

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

u = 0

The following points are undefined

u = 0

Combine undefined points with solutions:

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

Substitute back

u = cos (x)

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

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

cos (x) = \frac{3}{2} : x = \frac{π}{6} + 2 πn, x = \frac{11 π}{6} + 2 πn

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

General solutions for

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

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}{6} + 2 πn, x = \frac{11 π}{6} + 2 πn

x = \frac{π}{6} + 2 πn, x = \frac{11 π}{6} + 2 πn

cos (x) = - \frac{3}{2} : x = \frac{5 π}{6} + 2 πn, x = \frac{7 π}{6} + 2 πn

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

General solutions for

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

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{5 π}{6} + 2 πn, x = \frac{7 π}{6} + 2 πn

x = \frac{5 π}{6} + 2 πn, x = \frac{7 π}{6} + 2 πn

Combine all the solutions

x = \frac{π}{6} + 2 πn, x = \frac{11 π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn, x = \frac{7 π}{6} + 2 πn

Popular Examples

cos(A)=-1/2

cos (A) = - \frac{1}{2}

sin(x)=cos(4x)

sin (x) = cos (4 x)

cos(x)= 13/12

cos (x) = \frac{13}{12}

-2+cot(θ)=-3

- 2 + cot (θ) = - 3

sin^2(x)+sin(x)=0,0<= x<= 2pi

sin^{2} (x) + sin (x) = 0, 0 \leq x \leq 2 π