What is the general solution for 25sin^2(x)-10sin(x)+1=0 ?

The general solution for 25sin^2(x)-10sin(x)+1=0 is x=0.20135…+2pin,x=pi-0.20135…+2pin

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25sin^2(x)-10sin(x)+1=0

25 sin^{2} (x) - 10 sin (x) + 1 = 0

x = 0.20135 \dots + 2 πn, x = π - 0.20135 \dots + 2 πn

Degrees

x = 11.53695 \dots^{\circ} + 36 0^{\circ} n, x = 168.46304 \dots^{\circ} + 36 0^{\circ} n

Solution steps

25 sin^{2} (x) - 10 sin (x) + 1 = 0

Solve by substitution

25 sin^{2} (x) - 10 sin (x) + 1 = 0

Let:

sin (x) = u

25 u^{2} - 10 u + 1 = 0

25 u^{2} - 10 u + 1 = 0 : u = \frac{1}{5}

25 u^{2} - 10 u + 1 = 0

Solve with the quadratic formula

25 u^{2} - 10 u + 1 = 0

Quadratic Equation Formula:

For

a = 25, b = - 10, c = 1

u_{1, 2} = \frac{- ( - 10 ) \pm ( - 10 ) ^{2} - 4 \cdot 25 \cdot 1}{2 \cdot 25}

u_{1, 2} = \frac{- ( - 10 ) \pm ( - 10 ) ^{2} - 4 \cdot 25 \cdot 1}{2 \cdot 25}

(- 10)^{2} - 4 \cdot 25 \cdot 1 = 0

(- 10)^{2} - 4 \cdot 25 \cdot 1

Apply exponent rule:

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

n

is even

(- 10)^{2} = 1 0^{2}

= 1 0^{2} - 4 \cdot 25 \cdot 1

Multiply the numbers:

4 \cdot 25 \cdot 1 = 100

= 1 0^{2} - 100

1 0^{2} = 100

= 100 - 100

Subtract the numbers:

100 - 100 = 0

= 0

u_{1, 2} = \frac{- ( - 10 ) \pm 0}{2 \cdot 25}

u = \frac{- ( - 10 )}{2 \cdot 25}

\frac{- ( - 10 )}{2 \cdot 25} = \frac{1}{5}

\frac{- ( - 10 )}{2 \cdot 25}

Apply rule

- (- a) = a

= \frac{10}{2 \cdot 25}

Multiply the numbers:

2 \cdot 25 = 50

= \frac{10}{50}

Cancel the common factor:

10

= \frac{1}{5}

u = \frac{1}{5}

The solution to the quadratic equation is:

u = \frac{1}{5}

Substitute back

u = sin (x)

sin (x) = \frac{1}{5}

sin (x) = \frac{1}{5}

sin (x) = \frac{1}{5} : x = arcsin (\frac{1}{5}) + 2 πn, x = π - arcsin (\frac{1}{5}) + 2 πn

sin (x) = \frac{1}{5}

Apply trig inverse properties

sin (x) = \frac{1}{5}

General solutions for

sin (x) = \frac{1}{5}

sin (x) = a \Rightarrow x = arcsin (a) + 2 πn, x = π - arcsin (a) + 2 πn

x = arcsin (\frac{1}{5}) + 2 πn, x = π - arcsin (\frac{1}{5}) + 2 πn

x = arcsin (\frac{1}{5}) + 2 πn, x = π - arcsin (\frac{1}{5}) + 2 πn

Combine all the solutions

x = arcsin (\frac{1}{5}) + 2 πn, x = π - arcsin (\frac{1}{5}) + 2 πn

Show solutions in decimal form

x = 0.20135 \dots + 2 πn, x = π - 0.20135 \dots + 2 πn

4 - 4 sin (θ) = 2

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

cos (x) = \frac{84}{98}

cos (\frac{3 θ}{2}) = \frac{1}{2}

0 = 5 + 4 cos (x)

What is the general solution for 25sin^2(x)-10sin(x)+1=0 ?
The general solution for 25sin^2(x)-10sin(x)+1=0 is x=0.20135…+2pin,x=pi-0.20135…+2pin