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Popular Trigonometry >

2sin^2(x)=cos^3(x)tan(x)

Solution

2 sin^{2} (x) = cos^{3} (x) tan (x)

Solution

x = 2 πn, x = π + 2 πn, x = 0.42707 \dots + 2 πn, x = π - 0.42707 \dots + 2 πn

Degrees

x = 0^{\circ} + 36 0^{\circ} n, x = 18 0^{\circ} + 36 0^{\circ} n, x = 24.46980 \dots^{\circ} + 36 0^{\circ} n, x = 155.53019 \dots^{\circ} + 36 0^{\circ} n

Solution steps

2 sin^{2} (x) = cos^{3} (x) tan (x)

Subtract

cos^{3} (x) tan (x)

from both sides

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

Rewrite using trig identities

2 sin^{2} (x) - cos^{3} (x) tan (x)

Use the basic trigonometric identity:

tan (x) = \frac{sin ( x )}{cos ( x )}

= 2 sin^{2} (x) - cos^{3} (x) \frac{sin ( x )}{cos ( x )}

cos^{3} (x) \frac{sin ( x )}{cos ( x )} = cos^{2} (x) sin (x)

cos^{3} (x) \frac{sin ( x )}{cos ( x )}

Multiply fractions:

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

= \frac{sin ( x ) cos ^{3} ( x )}{cos ( x )}

Cancel the common factor:

cos (x)

= cos^{2} (x) sin (x)

= 2 sin^{2} (x) - cos^{2} (x) sin (x)

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

Factor

2 sin^{2} (x) - cos^{2} (x) sin (x) : sin (x) (2 sin (x) - cos^{2} (x))

2 sin^{2} (x) - cos^{2} (x) sin (x)

Apply exponent rule:

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

sin^{2} (x) = sin (x) sin (x)

= 2 sin (x) sin (x) - sin (x) cos^{2} (x)

Factor out common term

sin (x)

= sin (x) (2 sin (x) - cos^{2} (x))

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

Solving each part separately

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

sin (x) = 0 : x = 2 πn, x = π + 2 πn

sin (x) = 0

General solutions for

sin (x) = 0

sin (x)

periodicity table with

2 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} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

x = 0 + 2 πn, x = π + 2 πn

x = 0 + 2 πn, x = π + 2 πn

Solve

x = 0 + 2 πn : x = 2 πn

x = 0 + 2 πn

0 + 2 πn = 2 πn

x = 2 πn

x = 2 πn, x = π + 2 πn

2 sin (x) - cos^{2} (x) = 0 : x = arcsin (- 1 + 2) + 2 πn, x = π - arcsin (- 1 + 2) + 2 πn

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

Rewrite using trig identities

- cos^{2} (x) + 2 sin (x)

Use the Pythagorean identity:

cos^{2} (x) + sin^{2} (x) = 1

cos^{2} (x) = 1 - sin^{2} (x)

= - (1 - sin^{2} (x)) + 2 sin (x)

- (1 - sin^{2} (x)) : - 1 + sin^{2} (x)

- (1 - sin^{2} (x))

Distribute parentheses

= - (1) - (- sin^{2} (x))

Apply minus-plus rules

- (- a) = a, - (a) = - a

= - 1 + sin^{2} (x)

= - 1 + sin^{2} (x) + 2 sin (x)

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

Solve by substitution

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

Let:

sin (x) = u

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

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

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

Write in the standard form

a x^{2} + b x + c = 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) = 22

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

Apply rule

- (- a) = a

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

Multiply the numbers:

4 \cdot 1 \cdot 1 = 4

= 2^{2} + 4

2^{2} = 4

= 4 + 4

Add the numbers:

4 + 4 = 8

= 8

Prime factorization of

8 : 2^{3}

8

8

divides by

28 = 4 \cdot 2

= 2 \cdot 4

4

divides by

24 = 2 \cdot 2

= 2 \cdot 2 \cdot 2

2

is a prime number, therefore no further factorization is possible

= 2 \cdot 2 \cdot 2

= 2^{3}

= 2^{3}

Apply exponent rule:

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

= 2^{2} \cdot 2

Apply radical rule:

= 2 2^{2}

Apply radical rule:

2^{2} = 2

= 22

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

Separate the solutions

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

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

\frac{- 2 + 2 2}{2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{- 2 + 2 2}{2}

Factor

- 2 + 22 : 2 (- 1 + 2)

- 2 + 22

Rewrite as

= - 2 \cdot 1 + 22

Factor out common term

2

= 2 (- 1 + 2)

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

Divide the numbers:

\frac{2}{2} = 1

= - 1 + 2

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

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

Multiply the numbers:

2 \cdot 1 = 2

= \frac{- 2 - 2 2}{2}

Factor

- 2 - 22 : - 2 (1 + 2)

- 2 - 22

Rewrite as

= - 2 \cdot 1 - 22

Factor out common term

2

= - 2 (1 + 2)

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

Divide the numbers:

\frac{2}{2} = 1

= - (1 + 2)

Negate

- (1 + 2) = - 1 - 2

= - 1 - 2

The solutions to the quadratic equation are:

u = - 1 + 2, u = - 1 - 2

Substitute back

u = sin (x)

sin (x) = - 1 + 2, sin (x) = - 1 - 2

sin (x) = - 1 + 2, sin (x) = - 1 - 2

sin (x) = - 1 + 2 : x = arcsin (- 1 + 2) + 2 πn, x = π - arcsin (- 1 + 2) + 2 πn

sin (x) = - 1 + 2

Apply trig inverse properties

sin (x) = - 1 + 2

General solutions for

sin (x) = - 1 + 2

sin (x) = a \Rightarrow x = arcsin (a) + 2 πn, x = π - arcsin (a) + 2 πn

x = arcsin (- 1 + 2) + 2 πn, x = π - arcsin (- 1 + 2) + 2 πn

x = arcsin (- 1 + 2) + 2 πn, x = π - arcsin (- 1 + 2) + 2 πn

sin (x) = - 1 - 2 :

No Solution

sin (x) = - 1 - 2

- 1 \leq sin (x) \leq 1

No Solution

Combine all the solutions

x = arcsin (- 1 + 2) + 2 πn, x = π - arcsin (- 1 + 2) + 2 πn

Combine all the solutions

x = 2 πn, x = π + 2 πn, x = arcsin (- 1 + 2) + 2 πn, x = π - arcsin (- 1 + 2) + 2 πn

Show solutions in decimal form

x = 2 πn, x = π + 2 πn, x = 0.42707 \dots + 2 πn, x = π - 0.42707 \dots + 2 πn

Graph

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