What is the general solution for 2sin^2(x)-7cos(x)+2=0 ?

The general solution for 2sin^2(x)-7cos(x)+2=0 is x= pi/3+2pin,x=(5pi)/3+2pin

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2sin^2(x)-7cos(x)+2=0

Solution

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

Solution

x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

Degrees

x = 6 0^{\circ} + 36 0^{\circ} n, x = 30 0^{\circ} + 36 0^{\circ} n

Solution steps

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

Rewrite using trig identities

2 + 2 sin^{2} (x) - 7 cos (x)

Use the Pythagorean identity:

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

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

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

Simplify

2 + 2 (1 - cos^{2} (x)) - 7 cos (x) : - 2 cos^{2} (x) - 7 cos (x) + 4

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

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

Add the numbers:

2 + 2 = 4

= - 2 cos^{2} (x) - 7 cos (x) + 4

= - 2 cos^{2} (x) - 7 cos (x) + 4

4 - 2 cos^{2} (x) - 7 cos (x) = 0

Solve by substitution

4 - 2 cos^{2} (x) - 7 cos (x) = 0

Let:

cos (x) = u

4 - 2 u^{2} - 7 u = 0

4 - 2 u^{2} - 7 u = 0 : u = - 4, u = \frac{1}{2}

4 - 2 u^{2} - 7 u = 0

Write in the standard form

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

- 2 u^{2} - 7 u + 4 = 0

Solve with the quadratic formula

- 2 u^{2} - 7 u + 4 = 0

Quadratic Equation Formula:

For

a = - 2, b = - 7, c = 4

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

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

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

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

Apply rule

- (- a) = a

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

Apply exponent rule:

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

n

is even

(- 7)^{2} = 7^{2}

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

Multiply the numbers:

4 \cdot 2 \cdot 4 = 32

= 7^{2} + 32

7^{2} = 49

= 49 + 32

Add the numbers:

49 + 32 = 81

= 81

Factor the number:

81 = 9^{2}

= 9^{2}

Apply radical rule:

9^{2} = 9

= 9

u_{1, 2} = \frac{- ( - 7 ) \pm 9}{2 ( - 2 )}

Separate the solutions

u_{1} = \frac{- ( - 7 ) + 9}{2 ( - 2 )}, u_{2} = \frac{- ( - 7 ) - 9}{2 ( - 2 )}

u = \frac{- ( - 7 ) + 9}{2 ( - 2 )} : - 4

\frac{- ( - 7 ) + 9}{2 ( - 2 )}

Remove parentheses:

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

= \frac{7 + 9}{- 2 \cdot 2}

Add the numbers:

7 + 9 = 16

= \frac{16}{- 2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{16}{- 4}

Apply the fraction rule:

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

= - \frac{16}{4}

Divide the numbers:

\frac{16}{4} = 4

= - 4

u = \frac{- ( - 7 ) - 9}{2 ( - 2 )} : \frac{1}{2}

\frac{- ( - 7 ) - 9}{2 ( - 2 )}

Remove parentheses:

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

= \frac{7 - 9}{- 2 \cdot 2}

Subtract the numbers:

7 - 9 = - 2

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

Multiply the numbers:

2 \cdot 2 = 4

= \frac{- 2}{- 4}

Apply the fraction rule:

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

= \frac{2}{4}

Cancel the common factor:

2

= \frac{1}{2}

The solutions to the quadratic equation are:

u = - 4, u = \frac{1}{2}

Substitute back

u = cos (x)

cos (x) = - 4, cos (x) = \frac{1}{2}

cos (x) = - 4, cos (x) = \frac{1}{2}

cos (x) = - 4 :

No Solution

cos (x) = - 4

- 1 \leq cos (x) \leq 1

No Solution

cos (x) = \frac{1}{2} : x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

cos (x) = \frac{1}{2}

General solutions for

cos (x) = \frac{1}{2}

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 = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

Combine all the solutions

x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

Popular Examples

5cos(θ)=0

5 cos (θ) = 0

sin(2θ)=4cos(θ)

sin (2 θ) = 4 cos (θ)

tan(θ)= 4/4

tan (θ) = \frac{4}{4}

tan(θ)= 4/6

tan (θ) = \frac{4}{6}

tan(θ)= 4/7

tan (θ) = \frac{4}{7}

Frequently Asked Questions (FAQ)

What is the general solution for 2sin^2(x)-7cos(x)+2=0 ?
The general solution for 2sin^2(x)-7cos(x)+2=0 is x= pi/3+2pin,x=(5pi)/3+2pin