What is the general solution for 3sin^2(θ)+6=2sin^2(θ)+7 ?

The general solution for 3sin^2(θ)+6=2sin^2(θ)+7 is θ= pi/2+2pin,θ=(3pi)/2+2pin

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Popular >

3sin^2(θ)+6=2sin^2(θ)+7

Solution

3 sin^{2} (θ) + 6 = 2 sin^{2} (θ) + 7

Solution

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

Degrees

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

Solution steps

3 sin^{2} (θ) + 6 = 2 sin^{2} (θ) + 7

Solve by substitution

3 sin^{2} (θ) + 6 = 2 sin^{2} (θ) + 7

Let:

sin (θ) = u

3 u^{2} + 6 = 2 u^{2} + 7

3 u^{2} + 6 = 2 u^{2} + 7 : u = 1, u = - 1

3 u^{2} + 6 = 2 u^{2} + 7

Move

6

to the right side

3 u^{2} + 6 = 2 u^{2} + 7

Subtract

6

from both sides

3 u^{2} + 6 - 6 = 2 u^{2} + 7 - 6

Simplify

3 u^{2} = 2 u^{2} + 1

3 u^{2} = 2 u^{2} + 1

Move

2 u^{2}

to the left side

3 u^{2} = 2 u^{2} + 1

Subtract

2 u^{2}

from both sides

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

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 (θ)

sin (θ) = 1, sin (θ) = - 1

sin (θ) = 1, sin (θ) = - 1

sin (θ) = 1 : θ = \frac{π}{2} + 2 πn

sin (θ) = 1

General solutions for

sin (θ) = 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}

θ = \frac{π}{2} + 2 πn

θ = \frac{π}{2} + 2 πn

sin (θ) = - 1 : θ = \frac{3 π}{2} + 2 πn

sin (θ) = - 1

General solutions for

sin (θ) = - 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}

θ = \frac{3 π}{2} + 2 πn

θ = \frac{3 π}{2} + 2 πn

Combine all the solutions

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

Popular Examples

10sin^2(4x)=5

10 sin^{2} (4 x) = 5

cosh(x)=2

cosh (x) = 2

sin(2x)-sin(x)=0,0<= x<= 2pi

sin (2 x) - sin (x) = 0, 0 \leq x \leq 2 π

-2cos(3θ)=1

- 2 cos (3 θ) = 1

sec^2(x)+0.5tan(x)-1=0

sec^{2} (x) + 0.5 tan (x) - 1 = 0

Frequently Asked Questions (FAQ)

What is the general solution for 3sin^2(θ)+6=2sin^2(θ)+7 ?
The general solution for 3sin^2(θ)+6=2sin^2(θ)+7 is θ= pi/2+2pin,θ=(3pi)/2+2pin