What is the general solution for -3=tan^2(θ)-2-2tan(θ) ?

The general solution for -3=tan^2(θ)-2-2tan(θ) is θ= pi/4+pin

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Popular >

-3=tan^2(θ)-2-2tan(θ)

Solution

- 3 = tan^{2} (θ) - 2 - 2 tan (θ)

Solution

θ = \frac{π}{4} + πn

Degrees

θ = 4 5^{\circ} + 18 0^{\circ} n

Solution steps

- 3 = tan^{2} (θ) - 2 - 2 tan (θ)

Switch sides

tan^{2} (θ) - 2 - 2 tan (θ) = - 3

Solve by substitution

tan^{2} (θ) - 2 - 2 tan (θ) = - 3

Let:

tan (θ) = u

u^{2} - 2 - 2 u = - 3

u^{2} - 2 - 2 u = - 3 : u = 1

u^{2} - 2 - 2 u = - 3

Move

3

to the left side

u^{2} - 2 - 2 u = - 3

Add

3

to both sides

u^{2} - 2 - 2 u + 3 = - 3 + 3

Simplify

u^{2} - 2 u + 1 = 0

u^{2} - 2 u + 1 = 0

Solve with the quadratic formula

u^{2} - 2 u + 1 = 0

Quadratic Equation Formula:

For

a = 1, b = - 2, c = 1

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

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

(- 2)^{2} - 4 \cdot 1 \cdot 1 = 0

(- 2)^{2} - 4 \cdot 1 \cdot 1

Apply exponent rule:

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

n

is even

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

= 2^{2} - 4 \cdot 1 \cdot 1

Multiply the numbers:

4 \cdot 1 \cdot 1 = 4

= 2^{2} - 4

2^{2} = 4

= 4 - 4

Subtract the numbers:

4 - 4 = 0

= 0

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

u = \frac{- ( - 2 )}{2 \cdot 1}

\frac{- ( - 2 )}{2 \cdot 1} = 1

\frac{- ( - 2 )}{2 \cdot 1}

Apply rule

- (- a) = a

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

Multiply the numbers:

2 \cdot 1 = 2

= \frac{2}{2}

Apply rule

\frac{a}{a} = 1

= 1

u = 1

The solution to the quadratic equation is:

u = 1

Substitute back

u = tan (θ)

tan (θ) = 1

tan (θ) = 1

tan (θ) = 1 : θ = \frac{π}{4} + πn

tan (θ) = 1

General solutions for

tan (θ) = 1

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}

θ = \frac{π}{4} + πn

θ = \frac{π}{4} + πn

Combine all the solutions

θ = \frac{π}{4} + πn

Popular Examples

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- 2 cos^{2} (x) - 9 sin (x) - 3 = 0

sin(2x)=-(2sqrt(6))/7

sin (2 x) = - \frac{2 6}{7}

tan(6x)-3tan(3x)=0

tan (6 x) - 3 tan (3 x) = 0

sin(a)= 2/5

sin (a) = \frac{2}{5}

cos^2(θ)-cos(θ)-1=0

cos^{2} (θ) - cos (θ) - 1 = 0

Frequently Asked Questions (FAQ)

What is the general solution for -3=tan^2(θ)-2-2tan(θ) ?
The general solution for -3=tan^2(θ)-2-2tan(θ) is θ= pi/4+pin