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

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

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Popular >

2sin^2(x)+cos^2(x)=2

Solution

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

Solution

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

Degrees

x = 9 0^{\circ} + 36 0^{\circ} n, x = 27 0^{\circ} + 36 0^{\circ} n

Solution steps

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

Subtract

2

from both sides

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

Simplify

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

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

Add similar elements:

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

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

Add/Subtract the numbers:

- 2 + 1 = - 1

= sin^{2} (x) - 1

= sin^{2} (x) - 1

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

Solve by substitution

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

Let:

sin (x) = u

- 1 + u^{2} = 0

- 1 + u^{2} = 0 : u = 1, u = - 1

- 1 + u^{2} = 0

Move

1

to the right side

- 1 + u^{2} = 0

Add

1

to both sides

- 1 + u^{2} + 1 = 0 + 1

Simplify

u^{2} = 1

u^{2} = 1

For

x^{2} = f (a)

the solutions are

x = f (a), - f (a)

u = 1, u = - 1

1 = 1

1

Apply rule

1 = 1

= 1

- 1 = - 1

- 1

Apply rule

1 = 1

= - 1

u = 1, u = - 1

Substitute back

u = sin (x)

sin (x) = 1, sin (x) = - 1

sin (x) = 1, sin (x) = - 1

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) = - 1 : x = \frac{3 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{3 π}{2} + 2 πn

x = \frac{3 π}{2} + 2 πn

Combine all the solutions

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

Popular Examples

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

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

sin^2(x)+3cos(x)-1=0

sin^{2} (x) + 3 cos (x) - 1 = 0

cos((3x-7)/2)=0

cos (\frac{3 x - 7}{2}) = 0

3tan^2(x)= 8/(sin^2(x))

3 tan^{2} (x) = \frac{8}{sin ^{2} ( x )}

sin^2(x)+sin^{22}(x)+sin^{23}(x)=1

sin^{2} (x) + sin^{22} (x) + sin^{23} (x) = 1

Frequently Asked Questions (FAQ)

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