What is the general solution for 7tan^2(θ)-4tan(θ)=0,-90<= θ<= 90 ?

Q: What is the general solution for 7tan^2(θ)-4tan(θ)=0,-90<= θ<= 90 ?

The general solution for 7tan^2(θ)-4tan(θ)=0,-90<= θ<= 90 is θ=0.51914…,θ=0

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7tan^2(θ)-4tan(θ)=0,-90<= θ<= 90

Solution

7 tan^{2} (θ) - 4 tan (θ) = 0, - 9 0^{\circ} \leq θ \leq 9 0^{\circ}

Solution

θ = 0.51914 \dots, θ = 0

Solution steps

7 tan^{2} (θ) - 4 tan (θ) = 0, - 9 0^{\circ} \leq θ \leq 9 0^{\circ}

Solve by substitution

7 tan^{2} (θ) - 4 tan (θ) = 0

Let:

tan (θ) = u

7 u^{2} - 4 u = 0

7 u^{2} - 4 u = 0 : u = \frac{4}{7}, u = 0

7 u^{2} - 4 u = 0

Solve with the quadratic formula

7 u^{2} - 4 u = 0

Quadratic Equation Formula:

For

a = 7, b = - 4, c = 0

u_{1, 2} = \frac{- ( - 4 ) \pm ( - 4 ) ^{2} - 4 \cdot 7 \cdot 0}{2 \cdot 7}

u_{1, 2} = \frac{- ( - 4 ) \pm ( - 4 ) ^{2} - 4 \cdot 7 \cdot 0}{2 \cdot 7}

(- 4)^{2} - 4 \cdot 7 \cdot 0 = 4

(- 4)^{2} - 4 \cdot 7 \cdot 0

Apply exponent rule:

(- a)^{n} = a^{n},

n

is even

(- 4)^{2} = 4^{2}

= 4^{2} - 4 \cdot 7 \cdot 0

Apply rule

0 \cdot a = 0

= 4^{2} - 0

4^{2} - 0 = 4^{2}

= 4^{2}

Apply radical rule:

n a^{n} = a,

assuming

a \geq 0

= 4

u_{1, 2} = \frac{- ( - 4 ) \pm 4}{2 \cdot 7}

Separate the solutions

u_{1} = \frac{- ( - 4 ) + 4}{2 \cdot 7}, u_{2} = \frac{- ( - 4 ) - 4}{2 \cdot 7}

u = \frac{- ( - 4 ) + 4}{2 \cdot 7} : \frac{4}{7}

\frac{- ( - 4 ) + 4}{2 \cdot 7}

Apply rule

- (- a) = a

= \frac{4 + 4}{2 \cdot 7}

Add the numbers:

4 + 4 = 8

= \frac{8}{2 \cdot 7}

Multiply the numbers:

2 \cdot 7 = 14

= \frac{8}{14}

Cancel the common factor:

2

= \frac{4}{7}

u = \frac{- ( - 4 ) - 4}{2 \cdot 7} : 0

\frac{- ( - 4 ) - 4}{2 \cdot 7}

Apply rule

- (- a) = a

= \frac{4 - 4}{2 \cdot 7}

Subtract the numbers:

4 - 4 = 0

= \frac{0}{2 \cdot 7}

Multiply the numbers:

2 \cdot 7 = 14

= \frac{0}{14}

Apply rule

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

= 0

The solutions to the quadratic equation are:

u = \frac{4}{7}, u = 0

Substitute back

u = tan (θ)

tan (θ) = \frac{4}{7}, tan (θ) = 0

tan (θ) = \frac{4}{7}, tan (θ) = 0

tan (θ) = \frac{4}{7}, - 9 0^{\circ} \leq θ \leq 9 0^{\circ} : θ = arctan (\frac{4}{7})

tan (θ) = \frac{4}{7}, - 9 0^{\circ} \leq θ \leq 9 0^{\circ}

Apply trig inverse properties

tan (θ) = \frac{4}{7}

General solutions for

tan (θ) = \frac{4}{7}

tan (x) = a \Rightarrow x = arctan (a) + 18 0^{\circ} n

θ = arctan (\frac{4}{7}) + 18 0^{\circ} n

θ = arctan (\frac{4}{7}) + 18 0^{\circ} n

Solutions for the range

- 9 0^{\circ} \leq θ \leq 9 0^{\circ}

θ = arctan (\frac{4}{7})

tan (θ) = 0, - 9 0^{\circ} \leq θ \leq 9 0^{\circ} : θ = 0

tan (θ) = 0, - 9 0^{\circ} \leq θ \leq 9 0^{\circ}

General solutions for

tan (θ) = 0

tan (x)

periodicity table with

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

θ = 0 + 18 0^{\circ} n

θ = 0 + 18 0^{\circ} n

Solve

θ = 0 + 18 0^{\circ} n : θ = 18 0^{\circ} n

θ = 0 + 18 0^{\circ} n

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

θ = 18 0^{\circ} n

θ = 18 0^{\circ} n

Solutions for the range

- 9 0^{\circ} \leq θ \leq 9 0^{\circ}

θ = 0

Combine all the solutions

θ = arctan (\frac{4}{7}), θ = 0

Show solutions in decimal form

θ = 0.51914 \dots, θ = 0

Graph

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Frequently Asked Questions (FAQ)

What is the general solution for 7tan^2(θ)-4tan(θ)=0,-90<= θ<= 90 ?
The general solution for 7tan^2(θ)-4tan(θ)=0,-90<= θ<= 90 is θ=0.51914…,θ=0