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

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

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sin^2(x)=(1-cos^2(x))/2

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

x = 2 πn, x = π + 2 πn

Degrees

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

Solution steps

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

Subtract

\frac{1 - cos ^{2} ( x )}{2}

from both sides

sin^{2} (x) - \frac{1 - cos ^{2} ( x )}{2} = 0

Simplify

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

sin^{2} (x) - \frac{1 - cos ^{2} ( x )}{2}

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

Expand

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

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

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

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

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

Distribute parentheses

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

Apply minus-plus rules

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

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

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

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

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

\frac{f ( x )}{g ( x )} = 0 \Rightarrow f (x) = 0

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

Simplify

= sin^{2} (x)

sin^{2} (x) = 0

Apply rule

x^{n} = 0 \Rightarrow x = 0

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

cos (x) = - \frac{1}{2}, 0 < x < 2 π

cot (θ) = cos (θ)

3 sin^{2} (θ) - cos (θ) + 1 = 0

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

3 csc^{3} (x) - 4 csc (x) = 0

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