What is the general solution for sec^2(x)+0.5tan(x)-1=0 ?

The general solution for sec^2(x)+0.5tan(x)-1=0 is x=pin,x=-0.46364…+pin

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Popular >

sec^2(x)+0.5tan(x)-1=0

Solution

sec^{2} (x) + 0.5 tan (x) - 1 = 0

Solution

x = πn, x = - 0.46364 \dots + πn

Degrees

x = 0^{\circ} + 18 0^{\circ} n, x = - 26.56505 \dots^{\circ} + 18 0^{\circ} n

Solution steps

sec^{2} (x) + 0.5 tan (x) - 1 = 0

Rewrite using trig identities

- 1 + sec^{2} (x) + 0.5 tan (x)

Use the Pythagorean identity:

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

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

= 0.5 tan (x) + tan^{2} (x)

tan^{2} (x) + 0.5 tan (x) = 0

Solve by substitution

tan^{2} (x) + 0.5 tan (x) = 0

Let:

tan (x) = u

u^{2} + 0.5 u = 0

u^{2} + 0.5 u = 0 : u = 0, u = - \frac{1}{2}

u^{2} + 0.5 u = 0

Multiply both sides by

10

u^{2} + 0.5 u = 0

To eliminate decimal points, multiply by 10 for every digit after the decimal pointThere is one digit to the right of the decimal point, therefore multiply by

10

u^{2} \cdot 10 + 0.5 u \cdot 10 = 0 \cdot 10

Refine

10 u^{2} + 5 u = 0

10 u^{2} + 5 u = 0

Solve with the quadratic formula

10 u^{2} + 5 u = 0

Quadratic Equation Formula:

For

a = 10, b = 5, c = 0

u_{1, 2} = \frac{- 5 \pm 5 ^{2} - 4 \cdot 10 \cdot 0}{2 \cdot 10}

u_{1, 2} = \frac{- 5 \pm 5 ^{2} - 4 \cdot 10 \cdot 0}{2 \cdot 10}

5^{2} - 4 \cdot 10 \cdot 0 = 5

5^{2} - 4 \cdot 10 \cdot 0

Apply rule

0 \cdot a = 0

= 5^{2} - 0

5^{2} - 0 = 5^{2}

= 5^{2}

Apply radical rule:

assuming

a \geq 0

= 5

u_{1, 2} = \frac{- 5 \pm 5}{2 \cdot 10}

Separate the solutions

u_{1} = \frac{- 5 + 5}{2 \cdot 10}, u_{2} = \frac{- 5 - 5}{2 \cdot 10}

u = \frac{- 5 + 5}{2 \cdot 10} : 0

\frac{- 5 + 5}{2 \cdot 10}

Add/Subtract the numbers:

- 5 + 5 = 0

= \frac{0}{2 \cdot 10}

Multiply the numbers:

2 \cdot 10 = 20

= \frac{0}{20}

Apply rule

\frac{0}{a} = 0, a \neq = 0

= 0

u = \frac{- 5 - 5}{2 \cdot 10} : - \frac{1}{2}

\frac{- 5 - 5}{2 \cdot 10}

Subtract the numbers:

- 5 - 5 = - 10

= \frac{- 10}{2 \cdot 10}

Multiply the numbers:

2 \cdot 10 = 20

= \frac{- 10}{20}

Apply the fraction rule:

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

= - \frac{10}{20}

Cancel the common factor:

10

= - \frac{1}{2}

The solutions to the quadratic equation are:

u = 0, u = - \frac{1}{2}

Substitute back

u = tan (x)

tan (x) = 0, tan (x) = - \frac{1}{2}

tan (x) = 0, tan (x) = - \frac{1}{2}

tan (x) = 0 : x = πn

tan (x) = 0

General solutions for

tan (x) = 0

tan (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} tan (x) 0 \frac{3}{3} 13 \pm \infty - 3 - 1 - \frac{3}{3}

x = 0 + πn

x = 0 + πn

Solve

x = 0 + πn : x = πn

x = 0 + πn

0 + πn = πn

x = πn

x = πn

tan (x) = - \frac{1}{2} : x = arctan (- \frac{1}{2}) + πn

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

Apply trig inverse properties

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

General solutions for

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

tan (x) = - a \Rightarrow x = arctan (- a) + πn

x = arctan (- \frac{1}{2}) + πn

x = arctan (- \frac{1}{2}) + πn

Combine all the solutions

x = πn, x = arctan (- \frac{1}{2}) + πn

Show solutions in decimal form

x = πn, x = - 0.46364 \dots + πn

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tan (x - \frac{π}{2}) = 1

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5 cos (x) = 4 sin (x) + 4

6sec^2(x)tan(x)=8tan(x)

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csc (θ) = 0

Frequently Asked Questions (FAQ)

What is the general solution for sec^2(x)+0.5tan(x)-1=0 ?
The general solution for sec^2(x)+0.5tan(x)-1=0 is x=pin,x=-0.46364…+pin