QuestionSolve for $\alpha$ in the following equations: a. $2 \cdot \cos \alpha - 1 = 0$ b. $\alpha \sqrt{2} - 4 \sin \alpha = 0$ c. $\alpha \sin^2 \alpha - 3 \sin \alpha + 1 = 0$ d. $\sqrt{3} \cot^2 \alpha - (\sqrt{3} + 1) \cot \alpha + 1 = 0$ with $0^{\circ} \leq \alpha \leq 150^{\circ}$ .

Studdy Solution

STEP 1

Assumptions1. The given problems are trigonometric equations. . The solutions for $\alpha$ lie in the interval $[0^{\circ},150^{\circ}]$ .

STEP 2

Let's solve the first equation $2 \cdot \cos \alpha -1 =0$ .

STEP 3

Rearrange the equation to isolate $\cos \alpha$ .
$\cos \alpha = \frac{1}{2}$

STEP 4

Find the angle $\alpha$ whose cosine is $\frac{1}{2}$ within the given interval.
$\alpha = \cos^{-1}\left(\frac{1}{2}\right)$

STEP 5

Calculate the value of $\alpha$ .
$\alpha =60^{\circ},300^{\circ}$

STEP 6

Since $300^{\circ}$ is not in the interval $[0^{\circ},150^{\circ}]$ , the solution for the first equation is $\alpha =60^{\circ}$ .

STEP 7

Now, let's solve the second equation $\alpha \sqrt{2} -4 \sin \alpha =0$ .

STEP 8

Rearrange the equation to isolate $\sin \alpha$ .
$\sin \alpha = \frac{\alpha \sqrt{2}}{4}$

STEP 9

This equation cannot be solved analytically. We need to use numerical methods or graphing to find the solution for $\alpha$ .

STEP 10

Let's solve the third equation $\alpha \sin ^{2} \alpha -3 \sin \alpha + =0$ .

STEP 11

This equation is a quadratic in $\sin \alpha$ . Let's solve it by using the quadratic formula.
$\sin \alpha = \frac{3 \pm \sqrt{9 -4\alpha}}{\alpha}$

STEP 12

Again, this equation cannot be solved analytically. We need to use numerical methods or graphing to find the solution for $\alpha$ .

STEP 13

Finally, let's solve the fourth equation $\sqrt{3} \cot ^{2} \alpha - (\sqrt{3} +) \cot \alpha + =0$ .

STEP 14

This equation is a quadratic in $\cot \alpha$ . Let's solve it by using the quadratic formula.
$\cot \alpha = \frac{\sqrt{3} + \pm \sqrt{(\sqrt{3} +)^2 -4\sqrt{3}}}{2\sqrt{3}}$

STEP 15

Calculate the value of $\cot \alpha$ .
$\cot \alpha =, -\sqrt{3}$

STEP 16

Find the angle $\alpha$ whose cotangent is $$ and $-\sqrt{3}$ within the given interval.
$\alpha = \cot^{-}(), \cot^{-}(-\sqrt{3})$

STEP 17

Calculate the value of $\alpha$ .
$\alpha =45^{\circ},120^{\circ}$

STEP 18

So, the solutions for the four equations are $\alpha =60^{\circ}$ for the first equation, numerical solution for the second and third equations, and $\alpha =45^{\circ},120^{\circ}$ for the fourth equation.

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Studdy Solution

STEP 1

Assumptions1. The given problems are trigonometric equations. . The solutions for α\alphaα lie in the interval [0∘,150∘][0^{\circ},150^{\circ}][0∘,150∘].

STEP 2

Let's solve the first equation 2⋅cos⁡α−1=02 \cdot \cos \alpha -1 =02⋅cosα−1=0.

STEP 3

Rearrange the equation to isolate cos⁡α\cos \alphacosα.cos⁡α=12\cos \alpha = \frac{1}{2}cosα=21​

STEP 4

Find the angle α\alphaα whose cosine is 12\frac{1}{2}21​ within the given interval.α=cos⁡−1(12)\alpha = \cos^{-1}\left(\frac{1}{2}\right)α=cos−1(21​)

STEP 5

Calculate the value of α\alphaα.α=60∘,300∘\alpha =60^{\circ},300^{\circ}α=60∘,300∘

STEP 6

Since 300∘300^{\circ}300∘ is not in the interval [0∘,150∘][0^{\circ},150^{\circ}][0∘,150∘], the solution for the first equation is α=60∘\alpha =60^{\circ}α=60∘.

STEP 7

Now, let's solve the second equation α2−4sin⁡α=0\alpha \sqrt{2} -4 \sin \alpha =0α2​−4sinα=0.

STEP 8

Rearrange the equation to isolate sin⁡α\sin \alphasinα.sin⁡α=α24\sin \alpha = \frac{\alpha \sqrt{2}}{4}sinα=4α2​​

STEP 9

This equation cannot be solved analytically. We need to use numerical methods or graphing to find the solution for α\alphaα.

STEP 10

Let's solve the third equation αsin⁡2α−3sin⁡α+=0\alpha \sin ^{2} \alpha -3 \sin \alpha + =0αsin2α−3sinα+=0.

STEP 11

This equation is a quadratic in sin⁡α\sin \alphasinα. Let's solve it by using the quadratic formula.sin⁡α=3±9−4αα\sin \alpha = \frac{3 \pm \sqrt{9 -4\alpha}}{\alpha}sinα=α3±9−4α​​

STEP 12

Again, this equation cannot be solved analytically. We need to use numerical methods or graphing to find the solution for α\alphaα.

STEP 13

Finally, let's solve the fourth equation 3cot⁡2α−(3+)cot⁡α+=0\sqrt{3} \cot ^{2} \alpha - (\sqrt{3} +) \cot \alpha + =03​cot2α−(3​+)cotα+=0.

STEP 14

This equation is a quadratic in cot⁡α\cot \alphacotα. Let's solve it by using the quadratic formula.cot⁡α=3+±(3+)2−4323\cot \alpha = \frac{\sqrt{3} + \pm \sqrt{(\sqrt{3} +)^2 -4\sqrt{3}}}{2\sqrt{3}}cotα=23​3​+±(3​+)2−43​​​

STEP 15

Calculate the value of cot⁡α\cot \alphacotα.cot⁡α=,−3\cot \alpha =, -\sqrt{3}cotα=,−3​

STEP 16

Find the angle α\alphaα whose cotangent is $$ and $-\sqrt{3}$ within the given interval.α=cot⁡−(),cot⁡−(−3)\alpha = \cot^{-}(), \cot^{-}(-\sqrt{3})α=cot−(),cot−(−3​)

STEP 17

Calculate the value of α\alphaα.α=45∘,120∘\alpha =45^{\circ},120^{\circ}α=45∘,120∘

STEP 18

So, the solutions for the four equations are α=60∘\alpha =60^{\circ}α=60∘ for the first equation, numerical solution for the second and third equations, and α=45∘,120∘\alpha =45^{\circ},120^{\circ}α=45∘,120∘ for the fourth equation.

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Assumptions1. The given problems are trigonometric equations. . The solutions for $\alpha$ lie in the interval $[0^{\circ},150^{\circ}]$ .

Let's solve the first equation $2 \cdot \cos \alpha -1 =0$ .

Rearrange the equation to isolate $\cos \alpha$ .
$\cos \alpha = \frac{1}{2}$

Find the angle $\alpha$ whose cosine is $\frac{1}{2}$ within the given interval.
$\alpha = \cos^{-1}\left(\frac{1}{2}\right)$

Calculate the value of $\alpha$ .
$\alpha =60^{\circ},300^{\circ}$

Since $300^{\circ}$ is not in the interval $[0^{\circ},150^{\circ}]$ , the solution for the first equation is $\alpha =60^{\circ}$ .

Now, let's solve the second equation $\alpha \sqrt{2} -4 \sin \alpha =0$ .

Rearrange the equation to isolate $\sin \alpha$ .
$\sin \alpha = \frac{\alpha \sqrt{2}}{4}$

This equation cannot be solved analytically. We need to use numerical methods or graphing to find the solution for $\alpha$ .

Let's solve the third equation $\alpha \sin ^{2} \alpha -3 \sin \alpha + =0$ .

This equation is a quadratic in $\sin \alpha$ . Let's solve it by using the quadratic formula.
$\sin \alpha = \frac{3 \pm \sqrt{9 -4\alpha}}{\alpha}$

Again, this equation cannot be solved analytically. We need to use numerical methods or graphing to find the solution for $\alpha$ .

Finally, let's solve the fourth equation $\sqrt{3} \cot ^{2} \alpha - (\sqrt{3} +) \cot \alpha + =0$ .

This equation is a quadratic in $\cot \alpha$ . Let's solve it by using the quadratic formula.
$\cot \alpha = \frac{\sqrt{3} + \pm \sqrt{(\sqrt{3} +)^2 -4\sqrt{3}}}{2\sqrt{3}}$

Calculate the value of $\cot \alpha$ .
$\cot \alpha =, -\sqrt{3}$

Find the angle $\alpha$ whose cotangent is $$ and $-\sqrt{3}$ within the given interval.
$\alpha = \cot^{-}(), \cot^{-}(-\sqrt{3})$

Calculate the value of $\alpha$ .
$\alpha =45^{\circ},120^{\circ}$

So, the solutions for the four equations are $\alpha =60^{\circ}$ for the first equation, numerical solution for the second and third equations, and $\alpha =45^{\circ},120^{\circ}$ for the fourth equation.