What is the general solution for 2(sin(x))2-5cos(x)-4=0 ?

The general solution for 2(sin(x))2-5cos(x)-4=0 is x= pi/2+2pin,x=pi+0.22131…+2pin

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular Trigonometry >

2(sin(x))2-5cos(x)-4=0

Solution

2 (sin (x)) 2 - 5 cos (x) - 4 = 0

Solution

x = \frac{π}{2} + 2 πn, x = π + 0.22131 \dots + 2 πn

Degrees

x = 9 0^{\circ} + 36 0^{\circ} n, x = 192.68038 \dots^{\circ} + 36 0^{\circ} n

Solution steps

2 (sin (x)) \cdot 2 - 5 cos (x) - 4 = 0

Add

5 cos (x)

to both sides

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

Square both sides

(4 sin (x) - 4)^{2} = (5 cos (x))^{2}

Subtract

(5 cos (x))^{2}

from both sides

(4 sin (x) - 4)^{2} - 25 cos^{2} (x) = 0

Rewrite using trig identities

(- 4 + 4 sin (x))^{2} - 25 cos^{2} (x)

Use the Pythagorean identity:

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

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

= (- 4 + 4 sin (x))^{2} - 25 (1 - sin^{2} (x))

Simplify

(- 4 + 4 sin (x))^{2} - 25 (1 - sin^{2} (x)) : 41 sin^{2} (x) - 32 sin (x) - 9

(- 4 + 4 sin (x))^{2} - 25 (1 - sin^{2} (x))

(- 4 + 4 sin (x))^{2} : 16 - 32 sin (x) + 16 sin^{2} (x)

Apply Perfect Square Formula:

(a + b)^{2} = a^{2} + 2 ab + b^{2}

a = - 4, b = 4 sin (x)

= (- 4)^{2} + 2 (- 4) \cdot 4 sin (x) + (4 sin (x))^{2}

Simplify

(- 4)^{2} + 2 (- 4) \cdot 4 sin (x) + (4 sin (x))^{2} : 16 - 32 sin (x) + 16 sin^{2} (x)

(- 4)^{2} + 2 (- 4) \cdot 4 sin (x) + (4 sin (x))^{2}

Remove parentheses:

(- a) = - a

= (- 4)^{2} - 2 \cdot 4 \cdot 4 sin (x) + (4 sin (x))^{2}

(- 4)^{2} = 16

(- 4)^{2}

Apply exponent rule:

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

n

is even

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

= 4^{2}

4^{2} = 16

= 16

2 \cdot 4 \cdot 4 sin (x) = 32 sin (x)

2 \cdot 4 \cdot 4 sin (x)

Multiply the numbers:

2 \cdot 4 \cdot 4 = 32

= 32 sin (x)

(4 sin (x))^{2} = 16 sin^{2} (x)

(4 sin (x))^{2}

Apply exponent rule:

(a \cdot b)^{n} = a^{n} b^{n}

= 4^{2} sin^{2} (x)

4^{2} = 16

= 16 sin^{2} (x)

= 16 - 32 sin (x) + 16 sin^{2} (x)

= 16 - 32 sin (x) + 16 sin^{2} (x)

= 16 - 32 sin (x) + 16 sin^{2} (x) - 25 (1 - sin^{2} (x))

Expand

- 25 (1 - sin^{2} (x)) : - 25 + 25 sin^{2} (x)

- 25 (1 - sin^{2} (x))

Apply the distributive law:

a (b - c) = ab - a c

a = - 25, b = 1, c = sin^{2} (x)

= - 25 \cdot 1 - (- 25) sin^{2} (x)

Apply minus-plus rules

- (- a) = a

= - 25 \cdot 1 + 25 sin^{2} (x)

Multiply the numbers:

25 \cdot 1 = 25

= - 25 + 25 sin^{2} (x)

= 16 - 32 sin (x) + 16 sin^{2} (x) - 25 + 25 sin^{2} (x)

Simplify

16 - 32 sin (x) + 16 sin^{2} (x) - 25 + 25 sin^{2} (x) : 41 sin^{2} (x) - 32 sin (x) - 9

16 - 32 sin (x) + 16 sin^{2} (x) - 25 + 25 sin^{2} (x)

Group like terms

= - 32 sin (x) + 16 sin^{2} (x) + 25 sin^{2} (x) + 16 - 25

Add similar elements:

16 sin^{2} (x) + 25 sin^{2} (x) = 41 sin^{2} (x)

= - 32 sin (x) + 41 sin^{2} (x) + 16 - 25

Add/Subtract the numbers:

16 - 25 = - 9

= 41 sin^{2} (x) - 32 sin (x) - 9

= 41 sin^{2} (x) - 32 sin (x) - 9

= 41 sin^{2} (x) - 32 sin (x) - 9

- 9 - 32 sin (x) + 41 sin^{2} (x) = 0

Solve by substitution

- 9 - 32 sin (x) + 41 sin^{2} (x) = 0

Let:

sin (x) = u

- 9 - 32 u + 41 u^{2} = 0

- 9 - 32 u + 41 u^{2} = 0 : u = 1, u = - \frac{9}{41}

- 9 - 32 u + 41 u^{2} = 0

Write in the standard form

a x^{2} + b x + c = 0

41 u^{2} - 32 u - 9 = 0

Solve with the quadratic formula

41 u^{2} - 32 u - 9 = 0

Quadratic Equation Formula:

For

a = 41, b = - 32, c = - 9

u_{1, 2} = \frac{- ( - 32 ) \pm ( - 32 ) ^{2} - 4 \cdot 41 ( - 9 )}{2 \cdot 41}

u_{1, 2} = \frac{- ( - 32 ) \pm ( - 32 ) ^{2} - 4 \cdot 41 ( - 9 )}{2 \cdot 41}

(- 32)^{2} - 4 \cdot 41 (- 9) = 50

(- 32)^{2} - 4 \cdot 41 (- 9)

Apply rule

- (- a) = a

= (- 32)^{2} + 4 \cdot 41 \cdot 9

Apply exponent rule:

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

n

is even

(- 32)^{2} = 3 2^{2}

= 3 2^{2} + 4 \cdot 41 \cdot 9

Multiply the numbers:

4 \cdot 41 \cdot 9 = 1476

= 3 2^{2} + 1476

3 2^{2} = 1024

= 1024 + 1476

Add the numbers:

1024 + 1476 = 2500

= 2500

Factor the number:

2500 = 5 0^{2}

= 5 0^{2}

Apply radical rule:

n a^{n} = a

5 0^{2} = 50

= 50

u_{1, 2} = \frac{- ( - 32 ) \pm 50}{2 \cdot 41}

Separate the solutions

u_{1} = \frac{- ( - 32 ) + 50}{2 \cdot 41}, u_{2} = \frac{- ( - 32 ) - 50}{2 \cdot 41}

u = \frac{- ( - 32 ) + 50}{2 \cdot 41} : 1

\frac{- ( - 32 ) + 50}{2 \cdot 41}

Apply rule

- (- a) = a

= \frac{32 + 50}{2 \cdot 41}

Add the numbers:

32 + 50 = 82

= \frac{82}{2 \cdot 41}

Multiply the numbers:

2 \cdot 41 = 82

= \frac{82}{82}

Apply rule

\frac{a}{a} = 1

= 1

u = \frac{- ( - 32 ) - 50}{2 \cdot 41} : - \frac{9}{41}

\frac{- ( - 32 ) - 50}{2 \cdot 41}

Apply rule

- (- a) = a

= \frac{32 - 50}{2 \cdot 41}

Subtract the numbers:

32 - 50 = - 18

= \frac{- 18}{2 \cdot 41}

Multiply the numbers:

2 \cdot 41 = 82

= \frac{- 18}{82}

Apply the fraction rule:

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

= - \frac{18}{82}

Cancel the common factor:

2

= - \frac{9}{41}

The solutions to the quadratic equation are:

u = 1, u = - \frac{9}{41}

Substitute back

u = sin (x)

sin (x) = 1, sin (x) = - \frac{9}{41}

sin (x) = 1, sin (x) = - \frac{9}{41}

sin (x) = 1 : x = \frac{π}{2} + 2 πn

sin (x) = 1

General solutions for

sin (x) = 1

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 = \frac{π}{2} + 2 πn

x = \frac{π}{2} + 2 πn

sin (x) = - \frac{9}{41} : x = arcsin (- \frac{9}{41}) + 2 πn, x = π + arcsin (\frac{9}{41}) + 2 πn

sin (x) = - \frac{9}{41}

Apply trig inverse properties

sin (x) = - \frac{9}{41}

General solutions for

sin (x) = - \frac{9}{41}

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

x = arcsin (- \frac{9}{41}) + 2 πn, x = π + arcsin (\frac{9}{41}) + 2 πn

x = arcsin (- \frac{9}{41}) + 2 πn, x = π + arcsin (\frac{9}{41}) + 2 πn

Combine all the solutions

x = \frac{π}{2} + 2 πn, x = arcsin (- \frac{9}{41}) + 2 πn, x = π + arcsin (\frac{9}{41}) + 2 πn

Verify solutions by plugging them into the original equation

Check the solutions by plugging them into

2 sin (x) 2 - 5 cos (x) - 4 = 0

Remove the ones that don't agree with the equation.

Check the solution

\frac{π}{2} + 2 πn :

True

\frac{π}{2} + 2 πn

Plug in

n = 1

\frac{π}{2} + 2 π 1

For

2 sin (x) 2 - 5 cos (x) - 4 = 0

plug in

x = \frac{π}{2} + 2 π 1

2 sin (\frac{π}{2} + 2 π 1) \cdot 2 - 5 cos (\frac{π}{2} + 2 π 1) - 4 = 0

Refine

0 = 0

\Rightarrow True

Check the solution

arcsin (- \frac{9}{41}) + 2 πn :

False

arcsin (- \frac{9}{41}) + 2 πn

Plug in

n = 1

arcsin (- \frac{9}{41}) + 2 π 1

For

2 sin (x) 2 - 5 cos (x) - 4 = 0

plug in

x = arcsin (- \frac{9}{41}) + 2 π 1

2 sin (arcsin (- \frac{9}{41}) + 2 π 1) \cdot 2 - 5 cos (arcsin (- \frac{9}{41}) + 2 π 1) - 4 = 0

Refine

- 9.75609 \dots = 0

\Rightarrow False

Check the solution

π + arcsin (\frac{9}{41}) + 2 πn :

True

π + arcsin (\frac{9}{41}) + 2 πn

Plug in

n = 1

π + arcsin (\frac{9}{41}) + 2 π 1

For

2 sin (x) 2 - 5 cos (x) - 4 = 0

plug in

x = π + arcsin (\frac{9}{41}) + 2 π 1

2 sin (π + arcsin (\frac{9}{41}) + 2 π 1) \cdot 2 - 5 cos (π + arcsin (\frac{9}{41}) + 2 π 1) - 4 = 0

Refine

0 = 0

\Rightarrow True

x = \frac{π}{2} + 2 πn, x = π + arcsin (\frac{9}{41}) + 2 πn

Show solutions in decimal form

x = \frac{π}{2} + 2 πn, x = π + 0.22131 \dots + 2 πn

Graph

Popular Examples

cos(x/2+20)=sin(x)

0=cos^2(x)+sin^2(x)-2sin(x),0<= x<= 2pi

sin(θ)=-7/25 ,sin(2θ)

cos(θ)= 9/12

csc(x)+cot(x)=-1,0<= x<= 2pi

Frequently Asked Questions (FAQ)

What is the general solution for 2(sin(x))2-5cos(x)-4=0 ?
The general solution for 2(sin(x))2-5cos(x)-4=0 is x= pi/2+2pin,x=pi+0.22131…+2pin