Solved on Feb 06, 2024

Solve the trigonometric equation $16 \cos x + 10 = 18$ for $0 \leq x \leq 360^{\circ}$ .

STEP 1

Assumptions
1. We are solving the equation $16 \cos x + 10 = 18$ .
2. The solution for $x$ must be within the interval $0 \leq x \leq 360^{\circ}$ .

STEP 2

First, we need to isolate the cosine term. To do this, we subtract 10 from both sides of the equation.
$16 \cos x + 10 - 10 = 18 - 10$

STEP 3

Simplify the equation to find the expression for $\cos x$ .
$16 \cos x = 8$

STEP 4

Now, we divide both sides by 16 to solve for $\cos x$ .
$\frac{16 \cos x}{16} = \frac{8}{16}$

STEP 5

Simplify the fraction on the right-hand side of the equation.
$\cos x = \frac{1}{2}$

STEP 6

We now need to find the values of $x$ that satisfy the equation $\cos x = \frac{1}{2}$ within the interval $0 \leq x \leq 360^{\circ}$ .

STEP 7

Recall that $\cos x = \frac{1}{2}$ has solutions in the first and fourth quadrants because cosine is positive in these quadrants.

STEP 8

Find the principal value of $x$ for which $\cos x = \frac{1}{2}$ . This is the value of $x$ in the first quadrant.
$x = \cos^{-1}\left(\frac{1}{2}\right)$

STEP 9

Calculate the principal value using the unit circle or a calculator.
$x = 60^{\circ}$

STEP 10

Now find the value of $x$ in the fourth quadrant. Since cosine is symmetrical about the vertical axis, the angle in the fourth quadrant is $360^{\circ} - 60^{\circ}$ .

STEP 11

Calculate the angle in the fourth quadrant.
$x = 360^{\circ} - 60^{\circ}$

STEP 12

Simplify to find the second solution for $x$ .
$x = 300^{\circ}$

STEP 13

We have found the two solutions for $x$ within the interval $0 \leq x \leq 360^{\circ}$ that satisfy the equation $\cos x = \frac{1}{2}$ .
The solutions are $x = 60^{\circ}$ and $x = 300^{\circ}$ .

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Solve the trigonometric equation $16 \cos x + 10 = 18$ for $0 \leq x \leq 360^{\circ}$ .

STEP 1

Assumptions
1. We are solving the equation $16 \cos x + 10 = 18$ .
2. The solution for $x$ must be within the interval $0 \leq x \leq 360^{\circ}$ .

STEP 2

First, we need to isolate the cosine term. To do this, we subtract 10 from both sides of the equation.
$16 \cos x + 10 - 10 = 18 - 10$

STEP 3

Simplify the equation to find the expression for $\cos x$ .
$16 \cos x = 8$

STEP 4

Now, we divide both sides by 16 to solve for $\cos x$ .
$\frac{16 \cos x}{16} = \frac{8}{16}$

STEP 5

Simplify the fraction on the right-hand side of the equation.
$\cos x = \frac{1}{2}$

STEP 6

We now need to find the values of $x$ that satisfy the equation $\cos x = \frac{1}{2}$ within the interval $0 \leq x \leq 360^{\circ}$ .

STEP 7

Recall that $\cos x = \frac{1}{2}$ has solutions in the first and fourth quadrants because cosine is positive in these quadrants.

STEP 8

Find the principal value of $x$ for which $\cos x = \frac{1}{2}$ . This is the value of $x$ in the first quadrant.
$x = \cos^{-1}\left(\frac{1}{2}\right)$

STEP 9

Calculate the principal value using the unit circle or a calculator.
$x = 60^{\circ}$

STEP 10

Now find the value of $x$ in the fourth quadrant. Since cosine is symmetrical about the vertical axis, the angle in the fourth quadrant is $360^{\circ} - 60^{\circ}$ .

STEP 11

Calculate the angle in the fourth quadrant.
$x = 360^{\circ} - 60^{\circ}$

STEP 12

Simplify to find the second solution for $x$ .
$x = 300^{\circ}$

STEP 13

We have found the two solutions for $x$ within the interval $0 \leq x \leq 360^{\circ}$ that satisfy the equation $\cos x = \frac{1}{2}$ .
The solutions are $x = 60^{\circ}$ and $x = 300^{\circ}$ .

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Solve the trigonometric equation 16cos⁡x+10=1816 \cos x + 10 = 1816cosx+10=18 for 0≤x≤360∘0 \leq x \leq 360^{\circ}0≤x≤360∘.

STEP 1

Assumptions1. We are solving the equation 16cos⁡x+10=1816 \cos x + 10 = 1816cosx+10=18.2. The solution for xxx must be within the interval 0≤x≤360∘0 \leq x \leq 360^{\circ}0≤x≤360∘.

STEP 2

First, we need to isolate the cosine term. To do this, we subtract 10 from both sides of the equation.16cos⁡x+10−10=18−1016 \cos x + 10 - 10 = 18 - 1016cosx+10−10=18−10

STEP 3

Simplify the equation to find the expression for cos⁡x\cos xcosx.16cos⁡x=816 \cos x = 816cosx=8

STEP 4

Now, we divide both sides by 16 to solve for cos⁡x\cos xcosx.16cos⁡x16=816\frac{16 \cos x}{16} = \frac{8}{16}1616cosx​=168​

STEP 5

Simplify the fraction on the right-hand side of the equation.cos⁡x=12\cos x = \frac{1}{2}cosx=21​

STEP 6

We now need to find the values of xxx that satisfy the equation cos⁡x=12\cos x = \frac{1}{2}cosx=21​ within the interval 0≤x≤360∘0 \leq x \leq 360^{\circ}0≤x≤360∘.

STEP 7

Recall that cos⁡x=12\cos x = \frac{1}{2}cosx=21​ has solutions in the first and fourth quadrants because cosine is positive in these quadrants.

STEP 8

Find the principal value of xxx for which cos⁡x=12\cos x = \frac{1}{2}cosx=21​. This is the value of xxx in the first quadrant.x=cos⁡−1(12)x = \cos^{-1}\left(\frac{1}{2}\right)x=cos−1(21​)

STEP 9

Calculate the principal value using the unit circle or a calculator.x=60∘x = 60^{\circ}x=60∘

STEP 10

Now find the value of xxx in the fourth quadrant. Since cosine is symmetrical about the vertical axis, the angle in the fourth quadrant is 360∘−60∘360^{\circ} - 60^{\circ}360∘−60∘.

STEP 11

Calculate the angle in the fourth quadrant.x=360∘−60∘x = 360^{\circ} - 60^{\circ}x=360∘−60∘

STEP 12

Simplify to find the second solution for xxx.x=300∘x = 300^{\circ}x=300∘

STEP 13

We have found the two solutions for xxx within the interval 0≤x≤360∘0 \leq x \leq 360^{\circ}0≤x≤360∘ that satisfy the equation cos⁡x=12\cos x = \frac{1}{2}cosx=21​.The solutions are x=60∘x = 60^{\circ}x=60∘ and x=300∘x = 300^{\circ}x=300∘.

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Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

Solve the trigonometric equation $16 \cos x + 10 = 18$ for $0 \leq x \leq 360^{\circ}$ .

Assumptions
1. We are solving the equation $16 \cos x + 10 = 18$ .
2. The solution for $x$ must be within the interval $0 \leq x \leq 360^{\circ}$ .

First, we need to isolate the cosine term. To do this, we subtract 10 from both sides of the equation.
$16 \cos x + 10 - 10 = 18 - 10$

Simplify the equation to find the expression for $\cos x$ .
$16 \cos x = 8$

Now, we divide both sides by 16 to solve for $\cos x$ .
$\frac{16 \cos x}{16} = \frac{8}{16}$

Simplify the fraction on the right-hand side of the equation.
$\cos x = \frac{1}{2}$

We now need to find the values of $x$ that satisfy the equation $\cos x = \frac{1}{2}$ within the interval $0 \leq x \leq 360^{\circ}$ .

Recall that $\cos x = \frac{1}{2}$ has solutions in the first and fourth quadrants because cosine is positive in these quadrants.

Find the principal value of $x$ for which $\cos x = \frac{1}{2}$ . This is the value of $x$ in the first quadrant.
$x = \cos^{-1}\left(\frac{1}{2}\right)$

Calculate the principal value using the unit circle or a calculator.
$x = 60^{\circ}$

Now find the value of $x$ in the fourth quadrant. Since cosine is symmetrical about the vertical axis, the angle in the fourth quadrant is $360^{\circ} - 60^{\circ}$ .

Calculate the angle in the fourth quadrant.
$x = 360^{\circ} - 60^{\circ}$

Simplify to find the second solution for $x$ .
$x = 300^{\circ}$

We have found the two solutions for $x$ within the interval $0 \leq x \leq 360^{\circ}$ that satisfy the equation $\cos x = \frac{1}{2}$ .
The solutions are $x = 60^{\circ}$ and $x = 300^{\circ}$ .