What is the general solution for sec^2(x)+tan(x)=1,0<= x<= 2pi ?

Q: What is the general solution for sec^2(x)+tan(x)=1,0<= x<= 2pi ?

The general solution for sec^2(x)+tan(x)=1,0<= x<= 2pi is x=(3pi)/4 ,x=(7pi)/4

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sec^2(x)+tan(x)=1,0<= x<= 2pi

Solution

sec^{2} (x) + tan (x) = 1, 0 \leq x \leq 2 π

Solution

x = \frac{3 π}{4}, x = \frac{7 π}{4}

Degrees

x = 13 5^{\circ}, x = 31 5^{\circ}

Solution steps

sec^{2} (x) + tan (x) = 1, 0 \leq x \leq 2 π

Subtract

1

from both sides

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

Rewrite using trig identities

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

Use the Pythagorean identity:

sec^{2} (x) = tan^{2} (x) + 1

sec^{2} (x) - 1 = tan^{2} (x)

= tan (x) + tan^{2} (x)

Use the basic trigonometric identity:

tan (x) = \frac{1}{cot ( x )}

= \frac{1}{cot ( x )} + (\frac{1}{cot ( x )})^{2}

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

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

Apply exponent rule:

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

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

Apply rule

1^{a} = 1

1^{2} = 1

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

= \frac{1}{cot ( x )} + \frac{1}{cot ^{2} ( x )}

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

Solve by substitution

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

Let:

cot (x) = u

\frac{1}{u ^{2}} + \frac{1}{u} = 0

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

\frac{1}{u ^{2}} + \frac{1}{u} = 0

Multiply by LCM

\frac{1}{u ^{2}} + \frac{1}{u} = 0

Find Least Common Multiplier of

u^{2}, u : u^{2}

u^{2}, u

Lowest Common Multiplier (LCM)

Compute an expression comprised of factors that appear either in

u^{2}

u

= u^{2}

Multiply by LCM=

u^{2}

\frac{1}{u ^{2}} u^{2} + \frac{1}{u} u^{2} = 0 \cdot u^{2}

Simplify

\frac{1}{u ^{2}} u^{2} + \frac{1}{u} u^{2} = 0 \cdot u^{2}

Simplify

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

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

Multiply fractions:

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

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

Cancel the common factor:

u^{2}

= 1

Simplify

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

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

Multiply fractions:

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

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

Multiply:

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

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

Cancel the common factor:

u

= u

Simplify

0 \cdot u^{2} : 0

0 \cdot u^{2}

Apply rule

0 \cdot a = 0

= 0

1 + u = 0

1 + u = 0

1 + u = 0

Move

1

to the right side

1 + u = 0

Subtract

1

from both sides

1 + u - 1 = 0 - 1

Simplify

u = - 1

u = - 1

Verify Solutions

Find undefined (singularity) points:

u = 0

Take the denominator(s) of

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

and compare to zero

Solve

u^{2} = 0 : u = 0

u^{2} = 0

Apply rule

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

u = 0

u = 0

The following points are undefined

u = 0

Combine undefined points with solutions:

u = - 1

Substitute back

u = cot (x)

cot (x) = - 1

cot (x) = - 1

cot (x) = - 1, 0 \leq x \leq 2 π : x = \frac{3 π}{4}, x = \frac{7 π}{4}

cot (x) = - 1, 0 \leq x \leq 2 π

General solutions for

cot (x) = - 1

cot (x)

periodicity table with

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} cot (x) \mp \infty 31 \frac{3}{3} 0 - \frac{3}{3} - 1 - 3

x = \frac{3 π}{4} + πn

x = \frac{3 π}{4} + πn

Solutions for the range

0 \leq x \leq 2 π

x = \frac{3 π}{4}, x = \frac{7 π}{4}

Combine all the solutions

x = \frac{3 π}{4}, x = \frac{7 π}{4}

Graph

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Frequently Asked Questions (FAQ)

What is the general solution for sec^2(x)+tan(x)=1,0<= x<= 2pi ?
The general solution for sec^2(x)+tan(x)=1,0<= x<= 2pi is x=(3pi)/4 ,x=(7pi)/4