What is the general solution for cos(2x)=sqrt(1-sin^2(2x)),0<= x<= 2pi ?

Q: What is the general solution for cos(2x)=sqrt(1-sin^2(2x)),0<= x<= 2pi ?

The general solution for cos(2x)=sqrt(1-sin^2(2x)),0<= x<= 2pi is No Solution for x\in\mathbb{R}

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cos(2x)=sqrt(1-sin^2(2x)),0<= x<= 2pi

Solution

cos (2 x) = 1 - sin^{2} (2 x), 0 \leq x \leq 2 π

Solution

No Solution for x \in R

Solution steps

cos (2 x) = 1 - sin^{2} (2 x), 0 \leq x \leq 2 π

Subtract

1 - sin^{2} (2 x)

from both sides

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

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

Solve by substitution

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

Let:

cos (2 x) = u

u - u^{2} = 0

u - u^{2} = 0 : u \geq 0

u - u^{2} = 0

Remove square roots

u - u^{2} = 0

Subtract

u

from both sides

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

Simplify

- u^{2} = - u

Square both sides:

u^{2} = u^{2}

u - u^{2} = 0

(- u^{2})^{2} = (- u)^{2}

Expand

(- u^{2})^{2} : u^{2}

(- u^{2})^{2}

Apply exponent rule:

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

n

is even

(- u^{2})^{2} = (u^{2})^{2}

= (u^{2})^{2}

Apply radical rule:

a = a^{\frac{1}{2}}

= ((u^{2})^{\frac{1}{2}})^{2}

Apply exponent rule:

(a^{b})^{c} = a^{b c}

= (u^{2})^{\frac{1}{2} \cdot 2}

\frac{1}{2} \cdot 2 = 1

\frac{1}{2} \cdot 2

Multiply fractions:

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

= \frac{1 \cdot 2}{2}

Cancel the common factor:

2

= 1

= u^{2}

Expand

(- u)^{2} : u^{2}

(- u)^{2}

Apply exponent rule:

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

n

is even

(- u)^{2} = u^{2}

= u^{2}

u^{2} = u^{2}

u^{2} = u^{2}

u^{2} = u^{2}

Both sides are equal

True for all u

Verify Solutions:

u < 0

False

, u = 0

True

, u > 0

True

u - u^{2} = 0

Combine domain interval with solution interval:

True for all u

Find the function intervals:

u < 0, u = 0, u > 0

u - u^{2} = 0

Find the even roots arguments zeroes:

Solve

u^{2} = 0 : u = 0

u^{2} = 0

Apply rule

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

u = 0

u = 0

The intervals are defined around the zeroes:

u < 0, u = 0, u > 0

Combine intervals with domain

u < 0, u = 0, u > 0

Check the solutions by plugging them into

u - u^{2} = 0

Remove the ones that don't agree with the equation.

Plug

u < 0 : u - u^{2} \neq = 0 \Rightarrow

False

The solution is

u \geq 0

Substitute back

u = cos (2 x)

cos (2 x) \geq 0

cos (2 x) \geq 0

cos (2 x) = u \geq 0, 0 \leq x \leq 2 π :

No Solution

cos (2 x) = u \geq 0, 0 \leq x \leq 2 π

Apply trig inverse properties

cos (2 x) = u \geq 0

General solutions for

cos (2 x) = u \geq 0

cos (x) = a \Rightarrow x = arccos (a) + 2 πn, x = - arccos (a) + 2 πn

2 x = arccos (u \geq 0) + 2 πn, 2 x = - arccos (u \geq 0) + 2 πn

2 x = arccos (u \geq 0) + 2 πn, 2 x = - arccos (u \geq 0) + 2 πn

No Solution

Solutions for the range

0 \leq x \leq 2 π

No Solution

Combine all the solutions

No Solution for x \in R

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Frequently Asked Questions (FAQ)

What is the general solution for cos(2x)=sqrt(1-sin^2(2x)),0<= x<= 2pi ?
The general solution for cos(2x)=sqrt(1-sin^2(2x)),0<= x<= 2pi is No Solution for x\in\mathbb{R}