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

The general solution for 2cos^2(x)-sqrt(2)sin(x)=2 is x=(5pi)/4+2pin,x=(7pi)/4+2pin,x=2pin,x=pi+2pin

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Popular >

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

Solution

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

Solution

x = \frac{5 π}{4} + 2 πn, x = \frac{7 π}{4} + 2 πn, x = 2 πn, x = π + 2 πn

Degrees

x = 22 5^{\circ} + 36 0^{\circ} n, x = 31 5^{\circ} + 36 0^{\circ} n, x = 0^{\circ} + 36 0^{\circ} n, x = 18 0^{\circ} + 36 0^{\circ} n

Solution steps

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

Subtract

2

from both sides

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

Simplify

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

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

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

Expand

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

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

Apply the distributive law:

a (b - c) = ab - a c

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

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

Multiply the numbers:

2 \cdot 1 = 2

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

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

- 2 + 2 = 0

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

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

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

Solve by substitution

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

Let:

sin (x) = u

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

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

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

Write in the standard form

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

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

a = - 2, b = - 2, c = 0

u_{1, 2} = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 ( - 2 ) \cdot 0}{2 ( - 2 )}

u_{1, 2} = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 ( - 2 ) \cdot 0}{2 ( - 2 )}

(- 2)^{2} - 4 (- 2) \cdot 0 = 2

(- 2)^{2} - 4 (- 2) \cdot 0

Apply rule

- (- a) = a

= (- 2)^{2} + 4 \cdot 2 \cdot 0

(- 2)^{2} = 2

(- 2)^{2}

Apply exponent rule:

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

n

is even

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

= (2)^{2}

Apply radical rule:

a = a^{\frac{1}{2}}

= (2^{\frac{1}{2}})^{2}

Apply exponent rule:

(a^{b})^{c} = a^{b c}

= 2^{\frac{1}{2} \cdot 2}

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

\frac{1}{2} \cdot 2

Multiply fractions:

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

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

Cancel the common factor:

2

= 1

= 2

4 \cdot 2 \cdot 0 = 0

4 \cdot 2 \cdot 0

Apply rule

0 \cdot a = 0

= 0

= 2 + 0

Add the numbers:

2 + 0 = 2

= 2

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

Separate the solutions

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

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

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

Remove parentheses:

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

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

Add similar elements:

2 + 2 = 22

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

Multiply the numbers:

2 \cdot 2 = 4

= \frac{2 2}{- 4}

Apply the fraction rule:

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

= - \frac{2 2}{4}

Cancel the common factor:

2

= - \frac{2}{2}

u = \frac{- ( - 2 ) - 2}{2 ( - 2 )} : 0

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

Remove parentheses:

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

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

Add similar elements:

2 - 2 = 0

= \frac{0}{- 2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{0}{- 4}

Apply the fraction rule:

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

= - \frac{0}{4}

Apply rule

\frac{0}{a} = 0, a \neq = 0

= - 0

= 0

The solutions to the quadratic equation are:

u = - \frac{2}{2}, u = 0

Substitute back

u = sin (x)

sin (x) = - \frac{2}{2}, sin (x) = 0

sin (x) = - \frac{2}{2}, sin (x) = 0

sin (x) = - \frac{2}{2} : x = \frac{5 π}{4} + 2 πn, x = \frac{7 π}{4} + 2 πn

sin (x) = - \frac{2}{2}

General solutions for

sin (x) = - \frac{2}{2}

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{5 π}{4} + 2 πn, x = \frac{7 π}{4} + 2 πn

x = \frac{5 π}{4} + 2 πn, x = \frac{7 π}{4} + 2 πn

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

Combine all the solutions

x = \frac{5 π}{4} + 2 πn, x = \frac{7 π}{4} + 2 πn, x = 2 πn, x = π + 2 πn

Popular Examples

sin(a)=-5/13

sin (a) = - \frac{5}{13}

sin^2(x)=((8m-2))/9

sin^{2} (x) = \frac{( 8 m - 2 )}{9}

0.06=0.1(cos(4(t)-1.57))

0.06 = 0.1 (cos (4 (t) - 1.57))

sin(x)= 13/14

sin (x) = \frac{13}{14}

sin(x-pi/2)+sin(pi/2)=sin(x+pi/2)

sin (x - \frac{π}{2}) + sin (\frac{π}{2}) = sin (x + \frac{π}{2})

Frequently Asked Questions (FAQ)

What is the general solution for 2cos^2(x)-sqrt(2)sin(x)=2 ?
The general solution for 2cos^2(x)-sqrt(2)sin(x)=2 is x=(5pi)/4+2pin,x=(7pi)/4+2pin,x=2pin,x=pi+2pin