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3 years ago

Sin\theta +cos\theta =1

Mathematics

1 answer:

Mamont248 [21]3 years ago

4 0

Answer:

θ = 2 π n_1 + π/2 for n_1 element Z or θ = 2 π n_2 for n_2 element Z

Step-by-step explanation:

Solve for θ:

cos(θ) + sin(θ) = 1

cos(θ) + sin(θ) = sqrt(2) (cos(θ)/sqrt(2) + sin(θ)/sqrt(2)) = sqrt(2) (sin(π/4) cos(θ) + cos(π/4) sin(θ)) = sqrt(2) sin(θ + π/4):

sqrt(2) sin(θ + π/4) = 1

Divide both sides by sqrt(2):

sin(θ + π/4) = 1/sqrt(2)

Take the inverse sine of both sides:

θ + π/4 = 2 π n_1 + (3 π)/4 for n_1 element Z

or θ + π/4 = 2 π n_2 + π/4 for n_2 element Z

Subtract π/4 from both sides:

θ = 2 π n_1 + π/2 for n_1 element Z

or θ + π/4 = 2 π n_2 + π/4 for n_2 element Z

Subtract π/4 from both sides:

Answer: θ = 2 π n_1 + π/2 for n_1 element Z

or θ = 2 π n_2 for n_2 element Z

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3 years ago

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$\displaystyle x = \frac{\pi}{3} +k\, \pi$ or $\displaystyle x =- \frac{\pi}{3} +2\,k\, \pi$ , where $k$ is an integer.

There are three such angles between $0$ and $2\pi$ : $\displaystyle \frac{\pi}{3}$ , $\displaystyle \frac{2\, \pi}{3}$ , and $\displaystyle \frac{4\,\pi}{3}$ .

Step-by-step explanation:

By the double angle identity of sines:

$\sin(2\, x) = 2\, \sin x \cdot \cos x$ .

Rewrite the original equation with this identity:

$2\, (2\, \sin x \cdot \cos x) - 2\, \sin x + 2\sqrt{3}\, \cos x - \sqrt{3} = 0$ .

Note, that $2\, (2\, \sin x \cdot \cos x)$ and $(-2\, \sin x)$ share the common factor $(2\, \sin x)$ . On the other hand, $2\sqrt{3}\, \cos x$ and $(-\sqrt{3})$ share the common factor $\sqrt[3}$ . Combine these terms pairwise using the two common factors:

$(2\, \sin x) \cdot (2\, \cos x - 1) + \left(\sqrt{3}\right)\, (2\, \cos x - 1) = 0$ .

Note the new common factor $(2\, \cos x - 1)$ . Therefore:

$\left(2\, \sin x + \sqrt{3}\right) \cdot (2\, \cos x - 1) = 0$ .

This equation holds as long as either $\left(2\, \sin x + \sqrt{3}\right)$ or $(2\, \cos x - 1)$ is zero. Let $k$ be an integer. Accordingly:

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$\displaystyle \cos x = \frac{1}{2}$ , which corresponds to $\displaystyle x = \frac{\pi}{3} + 2\, k \, \pi$ and $\displaystyle x = -\frac{\pi}{3} + 2\, k \, \pi$ .

Any $x$ that fits into at least one of these patterns will satisfy the equation. These pattern can be further combined:

$\displaystyle x = \frac{\pi}{3} + k \, \pi$ (from $\displaystyle x = -\frac{2\,\pi}{3} + 2\, k\, \pi$ and $\displaystyle x = \frac{\pi}{3} + 2\, k \, \pi$ , combined,) as well as
$\displaystyle x =- \frac{\pi}{3} +2\,k\, \pi$ .