What is the general solution for 3sin^2(x)+2cos(2x)=1 ?

The general solution for 3sin^2(x)+2cos(2x)=1 is x= pi/2+2pin,x=(3pi)/2+2pin

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

Solution

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

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

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

Subtract

1

from both sides

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

Rewrite using trig identities

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

Use the Double Angle identity:

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

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

Simplify

- 1 + 2 (1 - 2 sin^{2} (x)) + 3 sin^{2} (x) : - sin^{2} (x) + 1

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

Expand

2 (1 - 2 sin^{2} (x)) : 2 - 4 sin^{2} (x)

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

Apply the distributive law:

a (b - c) = ab - a c

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

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

Simplify

2 \cdot 1 - 2 \cdot 2 sin^{2} (x) : 2 - 4 sin^{2} (x)

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

Multiply the numbers:

2 \cdot 1 = 2

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

Multiply the numbers:

2 \cdot 2 = 4

= 2 - 4 sin^{2} (x)

= 2 - 4 sin^{2} (x)

= - 1 + 2 - 4 sin^{2} (x) + 3 sin^{2} (x)

Simplify

- 1 + 2 - 4 sin^{2} (x) + 3 sin^{2} (x) : - sin^{2} (x) + 1

- 1 + 2 - 4 sin^{2} (x) + 3 sin^{2} (x)

Add similar elements:

- 4 sin^{2} (x) + 3 sin^{2} (x) = - sin^{2} (x)

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

Add/Subtract the numbers:

- 1 + 2 = 1

= - sin^{2} (x) + 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

Subtract

1

from both sides

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

Simplify

- u^{2} = - 1

- u^{2} = - 1

Divide both sides by

- 1

- u^{2} = - 1

Divide both sides by

- 1

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

1=sin(x)+sqrt(3)cos(x)

1 = sin (x) + 3 cos (x)

1+sin(x)=2cos^2(x),0<= x<5

1 + sin (x) = 2 cos^{2} (x), 0 \leq x < 5

2-3cos(x)=5+3cos(x)

2 - 3 cos (x) = 5 + 3 cos (x)

sinh(x)= 6/5

sinh (x) = \frac{6}{5}

tan(2x)tan(x)=1

tan (2 x) tan (x) = 1

Frequently Asked Questions (FAQ)

What is the general solution for 3sin^2(x)+2cos(2x)=1 ?
The general solution for 3sin^2(x)+2cos(2x)=1 is x= pi/2+2pin,x=(3pi)/2+2pin