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

The general solution for -(1/2)^2+cos(θ^2)=1 is No Solution for θ\in\mathbb{R}

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-(1/2)^2+cos(θ^2)=1

- (\frac{1}{2})^{2} + cos (θ^{2}) = 1

No Solution for θ \in R

Solution steps

- (\frac{1}{2})^{2} + cos (θ^{2}) = 1

Simplify

- (\frac{1}{2})^{2} + cos (θ^{2}) : - \frac{1}{4} + cos (θ^{2})

- (\frac{1}{2})^{2} + cos (θ^{2})

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

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

Apply exponent rule:

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

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

Apply rule

1^{a} = 1

1^{2} = 1

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

= - \frac{1}{2 ^{2}} + cos (θ^{2})

2^{2} = 4

= - \frac{1}{4} + cos (θ^{2})

- \frac{1}{4} + cos (θ^{2}) = 1

Move

\frac{1}{4}

to the right side

- \frac{1}{4} + cos (θ^{2}) = 1

Add

\frac{1}{4}

to both sides

- \frac{1}{4} + cos (θ^{2}) + \frac{1}{4} = 1 + \frac{1}{4}

Simplify

- \frac{1}{4} + cos (θ^{2}) + \frac{1}{4} = 1 + \frac{1}{4}

Simplify

- \frac{1}{4} + cos (θ^{2}) + \frac{1}{4} : cos (θ^{2})

- \frac{1}{4} + cos (θ^{2}) + \frac{1}{4}

Add similar elements:

- \frac{1}{4} + \frac{1}{4} = 0

= cos (θ^{2})

Simplify

1 + \frac{1}{4} : \frac{5}{4}

1 + \frac{1}{4}

Convert element to fraction:

1 = \frac{1 \cdot 4}{4}

= \frac{1 \cdot 4}{4} + \frac{1}{4}

Since the denominators are equal, combine the fractions:

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

= \frac{1 \cdot 4 + 1}{4}

1 \cdot 4 + 1 = 5

1 \cdot 4 + 1

Multiply the numbers:

1 \cdot 4 = 4

= 4 + 1

Add the numbers:

4 + 1 = 5

= 5

= \frac{5}{4}

cos (θ^{2}) = \frac{5}{4}

cos (θ^{2}) = \frac{5}{4}

cos (θ^{2}) = \frac{5}{4}

- 1 \leq cos (x) \leq 1

No Solution for θ \in R

0 = 2 cos (2 θ)

2 sec^{2} (x) + tan (x) = 8

tan (θ) = \frac{48}{8}

- 2 cos (2 x) = 2 sin (x)

sin (θ) - 0.2 cos (θ) = \frac{7.51}{9.8}

What is the general solution for -(1/2)^2+cos(θ^2)=1 ?
The general solution for -(1/2)^2+cos(θ^2)=1 is No Solution for θ\in\mathbb{R}