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

The general solution for 2sin^2(x)-cos(x)=0 is x=0.67488…+2pin,x=2pi-0.67488…+2pin

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

Solution

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

Solution

x = 0.67488 \dots + 2 πn, x = 2 π - 0.67488 \dots + 2 πn

Degrees

x = 38.66828 \dots^{\circ} + 36 0^{\circ} n, x = 321.33171 \dots^{\circ} + 36 0^{\circ} n

Solution steps

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

- cos (x) + (1 - cos^{2} (x)) \cdot 2 = 0

Solve by substitution

- cos (x) + (1 - cos^{2} (x)) \cdot 2 = 0

Let:

cos (x) = u

- u + (1 - u^{2}) \cdot 2 = 0

- u + (1 - u^{2}) \cdot 2 = 0 : u = - \frac{1 + 17}{4}, u = \frac{17 - 1}{4}

- u + (1 - u^{2}) \cdot 2 = 0

Expand

- u + (1 - u^{2}) \cdot 2 : - u + 2 - 2 u^{2}

- u + (1 - u^{2}) \cdot 2

= - u + 2 (1 - u^{2})

Expand

2 (1 - u^{2}) : 2 - 2 u^{2}

2 (1 - u^{2})

Apply the distributive law:

a (b - c) = ab - a c

a = 2, b = 1, c = u^{2}

= 2 \cdot 1 - 2 u^{2}

Multiply the numbers:

2 \cdot 1 = 2

= 2 - 2 u^{2}

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

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

Write in the standard form

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

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

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

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

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

(- 1)^{2} - 4 (- 2) \cdot 2 = 17

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

Apply rule

- (- a) = a

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

(- 1)^{2} = 1

(- 1)^{2}

Apply exponent rule:

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

n

is even

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

= 1^{2}

Apply rule

1^{a} = 1

= 1

4 \cdot 2 \cdot 2 = 16

4 \cdot 2 \cdot 2

Multiply the numbers:

4 \cdot 2 \cdot 2 = 16

= 16

= 1 + 16

Add the numbers:

1 + 16 = 17

= 17

u_{1, 2} = \frac{- ( - 1 ) \pm 17}{2 ( - 2 )}

Separate the solutions

u_{1} = \frac{- ( - 1 ) + 17}{2 ( - 2 )}, u_{2} = \frac{- ( - 1 ) - 17}{2 ( - 2 )}

u = \frac{- ( - 1 ) + 17}{2 ( - 2 )} : - \frac{1 + 17}{4}

\frac{- ( - 1 ) + 17}{2 ( - 2 )}

Remove parentheses:

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

= \frac{1 + 17}{- 2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{1 + 17}{- 4}

Apply the fraction rule:

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

= - \frac{1 + 17}{4}

u = \frac{- ( - 1 ) - 17}{2 ( - 2 )} : \frac{17 - 1}{4}

\frac{- ( - 1 ) - 17}{2 ( - 2 )}

Remove parentheses:

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

= \frac{1 - 17}{- 2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{1 - 17}{- 4}

Apply the fraction rule:

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

1 - 17 = - (17 - 1)

= \frac{17 - 1}{4}

The solutions to the quadratic equation are:

u = - \frac{1 + 17}{4}, u = \frac{17 - 1}{4}

Substitute back

u = cos (x)

cos (x) = - \frac{1 + 17}{4}, cos (x) = \frac{17 - 1}{4}

cos (x) = - \frac{1 + 17}{4}, cos (x) = \frac{17 - 1}{4}

cos (x) = - \frac{1 + 17}{4} :

No Solution

cos (x) = - \frac{1 + 17}{4}

- 1 \leq cos (x) \leq 1

No Solution

cos (x) = \frac{17 - 1}{4} : x = arccos (\frac{17 - 1}{4}) + 2 πn, x = 2 π - arccos (\frac{17 - 1}{4}) + 2 πn

cos (x) = \frac{17 - 1}{4}

Apply trig inverse properties

cos (x) = \frac{17 - 1}{4}

General solutions for

cos (x) = \frac{17 - 1}{4}

cos (x) = a \Rightarrow x = arccos (a) + 2 πn, x = 2 π - arccos (a) + 2 πn

x = arccos (\frac{17 - 1}{4}) + 2 πn, x = 2 π - arccos (\frac{17 - 1}{4}) + 2 πn

x = arccos (\frac{17 - 1}{4}) + 2 πn, x = 2 π - arccos (\frac{17 - 1}{4}) + 2 πn

Combine all the solutions

x = arccos (\frac{17 - 1}{4}) + 2 πn, x = 2 π - arccos (\frac{17 - 1}{4}) + 2 πn

Show solutions in decimal form

x = 0.67488 \dots + 2 πn, x = 2 π - 0.67488 \dots + 2 πn

Graph

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Frequently Asked Questions (FAQ)

What is the general solution for 2sin^2(x)-cos(x)=0 ?
The general solution for 2sin^2(x)-cos(x)=0 is x=0.67488…+2pin,x=2pi-0.67488…+2pin