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

The general solution for 2sin^2(x)-5cos(x)+5=0 is x=2pin

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Popular >

2sin^2(x)-5cos(x)+5=0

Solution

2 sin^{2} (x) - 5 cos (x) + 5 = 0

Solution

x = 2 πn

Degrees

x = 0^{\circ} + 36 0^{\circ} n

Solution steps

2 sin^{2} (x) - 5 cos (x) + 5 = 0

Rewrite using trig identities

5 + 2 sin^{2} (x) - 5 cos (x)

Use the Pythagorean identity:

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

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

= 5 + 2 (1 - cos^{2} (x)) - 5 cos (x)

Simplify

5 + 2 (1 - cos^{2} (x)) - 5 cos (x) : - 2 cos^{2} (x) - 5 cos (x) + 7

5 + 2 (1 - cos^{2} (x)) - 5 cos (x)

Expand

2 (1 - cos^{2} (x)) : 2 - 2 cos^{2} (x)

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

Apply the distributive law:

a (b - c) = ab - a c

a = 2, b = 1, c = cos^{2} (x)

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

Multiply the numbers:

2 \cdot 1 = 2

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

= 5 + 2 - 2 cos^{2} (x) - 5 cos (x)

Add the numbers:

5 + 2 = 7

= - 2 cos^{2} (x) - 5 cos (x) + 7

= - 2 cos^{2} (x) - 5 cos (x) + 7

7 - 2 cos^{2} (x) - 5 cos (x) = 0

Solve by substitution

7 - 2 cos^{2} (x) - 5 cos (x) = 0

Let:

cos (x) = u

7 - 2 u^{2} - 5 u = 0

7 - 2 u^{2} - 5 u = 0 : u = - \frac{7}{2}, u = 1

7 - 2 u^{2} - 5 u = 0

Write in the standard form

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

- 2 u^{2} - 5 u + 7 = 0

Solve with the quadratic formula

- 2 u^{2} - 5 u + 7 = 0

Quadratic Equation Formula:

For

a = - 2, b = - 5, c = 7

u_{1, 2} = \frac{- ( - 5 ) \pm ( - 5 ) ^{2} - 4 ( - 2 ) \cdot 7}{2 ( - 2 )}

u_{1, 2} = \frac{- ( - 5 ) \pm ( - 5 ) ^{2} - 4 ( - 2 ) \cdot 7}{2 ( - 2 )}

(- 5)^{2} - 4 (- 2) \cdot 7 = 9

(- 5)^{2} - 4 (- 2) \cdot 7

Apply rule

- (- a) = a

= (- 5)^{2} + 4 \cdot 2 \cdot 7

Apply exponent rule:

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

n

is even

(- 5)^{2} = 5^{2}

= 5^{2} + 4 \cdot 2 \cdot 7

Multiply the numbers:

4 \cdot 2 \cdot 7 = 56

= 5^{2} + 56

5^{2} = 25

= 25 + 56

Add the numbers:

25 + 56 = 81

= 81

Factor the number:

81 = 9^{2}

= 9^{2}

Apply radical rule:

9^{2} = 9

= 9

u_{1, 2} = \frac{- ( - 5 ) \pm 9}{2 ( - 2 )}

Separate the solutions

u_{1} = \frac{- ( - 5 ) + 9}{2 ( - 2 )}, u_{2} = \frac{- ( - 5 ) - 9}{2 ( - 2 )}

u = \frac{- ( - 5 ) + 9}{2 ( - 2 )} : - \frac{7}{2}

\frac{- ( - 5 ) + 9}{2 ( - 2 )}

Remove parentheses:

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

= \frac{5 + 9}{- 2 \cdot 2}

Add the numbers:

5 + 9 = 14

= \frac{14}{- 2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{14}{- 4}

Apply the fraction rule:

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

= - \frac{14}{4}

Cancel the common factor:

2

= - \frac{7}{2}

u = \frac{- ( - 5 ) - 9}{2 ( - 2 )} : 1

\frac{- ( - 5 ) - 9}{2 ( - 2 )}

Remove parentheses:

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

= \frac{5 - 9}{- 2 \cdot 2}

Subtract the numbers:

5 - 9 = - 4

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

Multiply the numbers:

2 \cdot 2 = 4

= \frac{- 4}{- 4}

Apply the fraction rule:

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

= \frac{4}{4}

Apply rule

\frac{a}{a} = 1

= 1

The solutions to the quadratic equation are:

u = - \frac{7}{2}, u = 1

Substitute back

u = cos (x)

cos (x) = - \frac{7}{2}, cos (x) = 1

cos (x) = - \frac{7}{2}, cos (x) = 1

cos (x) = - \frac{7}{2} :

No Solution

cos (x) = - \frac{7}{2}

- 1 \leq cos (x) \leq 1

No Solution

cos (x) = 1 : x = 2 πn

cos (x) = 1

General solutions for

cos (x) = 1

cos (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} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{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} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

x = 0 + 2 πn

x = 0 + 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

Combine all the solutions

x = 2 πn

Popular Examples

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tan (x) = \frac{20}{40}

sqrt(2)sin(x+pi/4)=0

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sin(x)=(2(5.5*10^{-7}))/(3.36*10^{-5)}

sin (x) = \frac{2 ( 5.5 \cdot 1 0 ^{- 7} )}{3.36 \cdot 1 0 ^{- 5}}

sin(x-30)=cos(15)

sin (x - 3 0^{\circ}) = cos (1 5^{\circ})

tan(x)=2.8867

tan (x) = 2.8867

Frequently Asked Questions (FAQ)

What is the general solution for 2sin^2(x)-5cos(x)+5=0 ?
The general solution for 2sin^2(x)-5cos(x)+5=0 is x=2pin