What is the general solution for tan(x)(tan(x)-3)+2=0 ?

The general solution for tan(x)(tan(x)-3)+2=0 is x=1.10714…+pin,x= pi/4+pin

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tan(x)(tan(x)-3)+2=0

Solution

tan (x) (tan (x) - 3) + 2 = 0

Solution

x = 1.10714 \dots + πn, x = \frac{π}{4} + πn

Degrees

x = 63.43494 \dots^{\circ} + 18 0^{\circ} n, x = 4 5^{\circ} + 18 0^{\circ} n

Solution steps

tan (x) (tan (x) - 3) + 2 = 0

Solve by substitution

tan (x) (tan (x) - 3) + 2 = 0

Let:

tan (x) = u

u (u - 3) + 2 = 0

u (u - 3) + 2 = 0 : u = 2, u = 1

u (u - 3) + 2 = 0

Expand

u (u - 3) + 2 : u^{2} - 3 u + 2

u (u - 3) + 2

Expand

u (u - 3) : u^{2} - 3 u

u (u - 3)

Apply the distributive law:

a (b - c) = ab - a c

a = u, b = u, c = 3

= uu - u \cdot 3

= uu - 3 u

uu = u^{2}

uu

Apply exponent rule:

a^{b} \cdot a^{c} = a^{b + c}

uu = u^{1 + 1}

= u^{1 + 1}

Add the numbers:

1 + 1 = 2

= u^{2}

= u^{2} - 3 u

= u^{2} - 3 u + 2

u^{2} - 3 u + 2 = 0

Solve with the quadratic formula

u^{2} - 3 u + 2 = 0

Quadratic Equation Formula:

For

a = 1, b = - 3, c = 2

u_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 \cdot 1 \cdot 2}{2 \cdot 1}

u_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 \cdot 1 \cdot 2}{2 \cdot 1}

(- 3)^{2} - 4 \cdot 1 \cdot 2 = 1

(- 3)^{2} - 4 \cdot 1 \cdot 2

Apply exponent rule:

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

n

is even

(- 3)^{2} = 3^{2}

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

Multiply the numbers:

4 \cdot 1 \cdot 2 = 8

= 3^{2} - 8

3^{2} = 9

= 9 - 8

Subtract the numbers:

9 - 8 = 1

= 1

Apply rule

1 = 1

= 1

u_{1, 2} = \frac{- ( - 3 ) \pm 1}{2 \cdot 1}

Separate the solutions

u_{1} = \frac{- ( - 3 ) + 1}{2 \cdot 1}, u_{2} = \frac{- ( - 3 ) - 1}{2 \cdot 1}

u = \frac{- ( - 3 ) + 1}{2 \cdot 1} : 2

\frac{- ( - 3 ) + 1}{2 \cdot 1}

Apply rule

- (- a) = a

= \frac{3 + 1}{2 \cdot 1}

Add the numbers:

3 + 1 = 4

= \frac{4}{2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{4}{2}

Divide the numbers:

\frac{4}{2} = 2

= 2

u = \frac{- ( - 3 ) - 1}{2 \cdot 1} : 1

\frac{- ( - 3 ) - 1}{2 \cdot 1}

Apply rule

- (- a) = a

= \frac{3 - 1}{2 \cdot 1}

Subtract the numbers:

3 - 1 = 2

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

Multiply the numbers:

2 \cdot 1 = 2

= \frac{2}{2}

Apply rule

\frac{a}{a} = 1

= 1

The solutions to the quadratic equation are:

u = 2, u = 1

Substitute back

u = tan (x)

tan (x) = 2, tan (x) = 1

tan (x) = 2, tan (x) = 1

tan (x) = 2 : x = arctan (2) + πn

tan (x) = 2

Apply trig inverse properties

tan (x) = 2

General solutions for

tan (x) = 2

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

x = arctan (2) + πn

x = arctan (2) + πn

tan (x) = 1 : x = \frac{π}{4} + πn

tan (x) = 1

General solutions for

tan (x) = 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}

x = \frac{π}{4} + πn

x = \frac{π}{4} + πn

Combine all the solutions

x = arctan (2) + πn, x = \frac{π}{4} + πn

Show solutions in decimal form

x = 1.10714 \dots + πn, x = \frac{π}{4} + πn

Popular Examples

sin(x)-4sin^3(x)=0

sin (x) - 4 sin^{3} (x) = 0

2sin(x)=3sqrt(3)

2 sin (x) = 33

sin^2(θ)= 1/9

sin^{2} (θ) = \frac{1}{9}

(sec(B)+csc(B))/(1+tan(B))=csc^2(B)

\frac{sec ( B ) + csc ( B )}{1 + tan ( B )} = csc^{2} (B)

4=sqrt(2)csc(3x)

4 = 2 csc (3 x)

Frequently Asked Questions (FAQ)

What is the general solution for tan(x)(tan(x)-3)+2=0 ?
The general solution for tan(x)(tan(x)-3)+2=0 is x=1.10714…+pin,x= pi/4+pin