Solved on Jan 12, 2024

Solve $4 \cos^3(2x) + 2 \cos^2(2x) - 2 \cos(2x) - 1 = 0$ for $x$ in $[0, 360^\circ)$ .

STEP 1

Assumptions
1. The equation to solve is $4 \cos^{3}(2x) + 2 \cos^{2}(2x) - 2 \cos(2x) - 1 = 0$ .
2. The domain for $x$ is $[0, 360^{\circ})$ .
3. The solutions for $x$ must be within the given domain.

STEP 2

Let $u = \cos(2x)$ . The equation becomes a cubic equation in terms of $u$ .
$4u^3 + 2u^2 - 2u - 1 = 0$

STEP 3

Now, we need to solve the cubic equation for $u$ . First, check if there are any obvious roots by substituting simple values such as $u = 0, \pm1$ .

STEP 4

Substitute $u = 1$ into the equation to check if it is a root.
$4(1)^3 + 2(1)^2 - 2(1) - 1 = 4 + 2 - 2 - 1 = 3 \neq 0$

STEP 5

Substitute $u = -1$ into the equation to check if it is a root.
$4(-1)^3 + 2(-1)^2 - 2(-1) - 1 = -4 + 2 + 2 - 1 = -1 \neq 0$

STEP 6

Since neither $u = 1$ nor $u = -1$ is a root, we must use other methods to find the roots. We can attempt to factor by grouping or use the Rational Root Theorem to find potential rational roots.

STEP 7

Try to factor the cubic equation by grouping terms or by synthetic division if a rational root is found.

STEP 8

Notice that $u = -1$ almost satisfies the equation, suggesting that $(u + 1)$ might be a factor. Let's rewrite the equation as:
$4u^3 + 4u^2 - 2u^2 - 2u + 2u - 1 = 0$

STEP 9

Now, factor by grouping:
$(4u^3 + 4u^2) + (-2u^2 - 2u) + (2u - 1) = 0$

STEP 10

Factor out the common terms in each group:
$4u^2(u + 1) - 2u(u + 1) + (2u - 1) = 0$

STEP 11

Notice that $4u^2(u + 1) - 2u(u + 1)$ can be factored further:
$u(4u^2 - 2)(u + 1) + (2u - 1) = 0$

STEP 12

Now, we have $u(4u - 2)(u + 1) + (2u - 1) = 0$ . We can see that $(2u - 1)$ is a common factor if we rewrite $4u^2 - 2$ as $2(2u - 1)$ :
$u(2(2u - 1))(u + 1) + (2u - 1) = 0$

STEP 13

Factor out $(2u - 1)$ :
$(2u - 1)(2u^2(u + 1) + 1) = 0$

STEP 14

Now, we have two factors to consider: $(2u - 1)$ and $(2u^2(u + 1) + 1)$ . Set each factor equal to zero and solve for $u$ .

STEP 15

First, solve $2u - 1 = 0$ for $u$ :
$2u - 1 = 0 \Rightarrow u = \frac{1}{2}$

STEP 16

Now, solve $2u^2(u + 1) + 1 = 0$ for $u$ . This is a quadratic in terms of $u^2$ .
$2u^2(u + 1) + 1 = 0 \Rightarrow 2u^3 + 2u^2 + 1 = 0$

STEP 17

This equation does not factor nicely, and it does not have rational roots. We can use the cubic formula or numerical methods to find the roots, but they will not be real since the discriminant is negative. Thus, the only real solution for $u$ is $u = \frac{1}{2}$ .

STEP 18

Now, recall that $u = \cos(2x)$ . Substitute $u = \frac{1}{2}$ back into this equation to solve for $x$ .
$\cos(2x) = \frac{1}{2}$

STEP 19

Solve for $2x$ by finding the angles whose cosine is $\frac{1}{2}$ . These angles are $60^{\circ}$ and $300^{\circ}$ .
$2x = 60^{\circ} \quad \text{or} \quad 2x = 300^{\circ}$

STEP 20

Divide each angle by 2 to solve for $x$ .
$x = \frac{60^{\circ}}{2} = 30^{\circ} \quad \text{or} \quad x = \frac{300^{\circ}}{2} = 150^{\circ}$

STEP 21

Check that the solutions are within the domain $[0, 360^{\circ})$ . Both $30^{\circ}$ and $150^{\circ}$ are within the domain.
The solutions for $x$ over the domain $[0, 360^{\circ})$ are $x = 30^{\circ}$ and $x = 150^{\circ}$ .

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Solve 4cos⁡3(2x)+2cos⁡2(2x)−2cos⁡(2x)−1=04 \cos^3(2x) + 2 \cos^2(2x) - 2 \cos(2x) - 1 = 04cos3(2x)+2cos2(2x)−2cos(2x)−1=0 for xxx in [0,360∘)[0, 360^\circ)[0,360∘).

STEP 1

Assumptions1. The equation to solve is 4cos⁡3(2x)+2cos⁡2(2x)−2cos⁡(2x)−1=04 \cos^{3}(2x) + 2 \cos^{2}(2x) - 2 \cos(2x) - 1 = 04cos3(2x)+2cos2(2x)−2cos(2x)−1=0.2. The domain for xxx is [0,360∘)[0, 360^{\circ})[0,360∘).3. The solutions for xxx must be within the given domain.

STEP 2

Let u=cos⁡(2x)u = \cos(2x)u=cos(2x). The equation becomes a cubic equation in terms of uuu.4u3+2u2−2u−1=04u^3 + 2u^2 - 2u - 1 = 04u3+2u2−2u−1=0

STEP 3

Now, we need to solve the cubic equation for uuu. First, check if there are any obvious roots by substituting simple values such as u=0,±1u = 0, \pm1u=0,±1.

STEP 4

Substitute u=1u = 1u=1 into the equation to check if it is a root.4(1)3+2(1)2−2(1)−1=4+2−2−1=3≠04(1)^3 + 2(1)^2 - 2(1) - 1 = 4 + 2 - 2 - 1 = 3 \neq 04(1)3+2(1)2−2(1)−1=4+2−2−1=3=0

STEP 5

Substitute u=−1u = -1u=−1 into the equation to check if it is a root.4(−1)3+2(−1)2−2(−1)−1=−4+2+2−1=−1≠04(-1)^3 + 2(-1)^2 - 2(-1) - 1 = -4 + 2 + 2 - 1 = -1 \neq 04(−1)3+2(−1)2−2(−1)−1=−4+2+2−1=−1=0

STEP 6

Since neither u=1u = 1u=1 nor u=−1u = -1u=−1 is a root, we must use other methods to find the roots. We can attempt to factor by grouping or use the Rational Root Theorem to find potential rational roots.

STEP 7

Try to factor the cubic equation by grouping terms or by synthetic division if a rational root is found.

STEP 8

Notice that u=−1u = -1u=−1 almost satisfies the equation, suggesting that (u+1)(u + 1)(u+1) might be a factor. Let's rewrite the equation as:4u3+4u2−2u2−2u+2u−1=04u^3 + 4u^2 - 2u^2 - 2u + 2u - 1 = 04u3+4u2−2u2−2u+2u−1=0

STEP 9

Now, factor by grouping:(4u3+4u2)+(−2u2−2u)+(2u−1)=0(4u^3 + 4u^2) + (-2u^2 - 2u) + (2u - 1) = 0(4u3+4u2)+(−2u2−2u)+(2u−1)=0

STEP 10

Factor out the common terms in each group:4u2(u+1)−2u(u+1)+(2u−1)=04u^2(u + 1) - 2u(u + 1) + (2u - 1) = 04u2(u+1)−2u(u+1)+(2u−1)=0

STEP 11

Notice that 4u2(u+1)−2u(u+1)4u^2(u + 1) - 2u(u + 1)4u2(u+1)−2u(u+1) can be factored further:u(4u2−2)(u+1)+(2u−1)=0u(4u^2 - 2)(u + 1) + (2u - 1) = 0u(4u2−2)(u+1)+(2u−1)=0

STEP 12

STEP 13

Factor out (2u−1)(2u - 1)(2u−1):(2u−1)(2u2(u+1)+1)=0(2u - 1)(2u^2(u + 1) + 1) = 0(2u−1)(2u2(u+1)+1)=0

STEP 14

Now, we have two factors to consider: (2u−1)(2u - 1)(2u−1) and (2u2(u+1)+1)(2u^2(u + 1) + 1)(2u2(u+1)+1). Set each factor equal to zero and solve for uuu.

STEP 15

First, solve 2u−1=02u - 1 = 02u−1=0 for uuu:2u−1=0⇒u=122u - 1 = 0 \Rightarrow u = \frac{1}{2}2u−1=0⇒u=21​

STEP 16

Now, solve 2u2(u+1)+1=02u^2(u + 1) + 1 = 02u2(u+1)+1=0 for uuu. This is a quadratic in terms of u2u^2u2.2u2(u+1)+1=0⇒2u3+2u2+1=02u^2(u + 1) + 1 = 0 \Rightarrow 2u^3 + 2u^2 + 1 = 02u2(u+1)+1=0⇒2u3+2u2+1=0

STEP 17

This equation does not factor nicely, and it does not have rational roots. We can use the cubic formula or numerical methods to find the roots, but they will not be real since the discriminant is negative. Thus, the only real solution for uuu is u=12u = \frac{1}{2}u=21​.

STEP 18

Now, recall that u=cos⁡(2x)u = \cos(2x)u=cos(2x). Substitute u=12u = \frac{1}{2}u=21​ back into this equation to solve for xxx.cos⁡(2x)=12\cos(2x) = \frac{1}{2}cos(2x)=21​

STEP 19

Solve for 2x2x2x by finding the angles whose cosine is 12\frac{1}{2}21​. These angles are 60∘60^{\circ}60∘ and 300∘300^{\circ}300∘.2x=60∘or2x=300∘2x = 60^{\circ} \quad \text{or} \quad 2x = 300^{\circ}2x=60∘or2x=300∘

STEP 20

Divide each angle by 2 to solve for xxx.x=60∘2=30∘orx=300∘2=150∘x = \frac{60^{\circ}}{2} = 30^{\circ} \quad \text{or} \quad x = \frac{300^{\circ}}{2} = 150^{\circ}x=260∘​=30∘orx=2300∘​=150∘

STEP 21

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Solve $4 \cos^3(2x) + 2 \cos^2(2x) - 2 \cos(2x) - 1 = 0$ for $x$ in $[0, 360^\circ)$ .

Assumptions
1. The equation to solve is $4 \cos^{3}(2x) + 2 \cos^{2}(2x) - 2 \cos(2x) - 1 = 0$ .
2. The domain for $x$ is $[0, 360^{\circ})$ .
3. The solutions for $x$ must be within the given domain.

Let $u = \cos(2x)$ . The equation becomes a cubic equation in terms of $u$ .
$4u^3 + 2u^2 - 2u - 1 = 0$

Now, we need to solve the cubic equation for $u$ . First, check if there are any obvious roots by substituting simple values such as $u = 0, \pm1$ .

Substitute $u = 1$ into the equation to check if it is a root.
$4(1)^3 + 2(1)^2 - 2(1) - 1 = 4 + 2 - 2 - 1 = 3 \neq 0$

Substitute $u = -1$ into the equation to check if it is a root.
$4(-1)^3 + 2(-1)^2 - 2(-1) - 1 = -4 + 2 + 2 - 1 = -1 \neq 0$

Since neither $u = 1$ nor $u = -1$ is a root, we must use other methods to find the roots. We can attempt to factor by grouping or use the Rational Root Theorem to find potential rational roots.

Notice that $u = -1$ almost satisfies the equation, suggesting that $(u + 1)$ might be a factor. Let's rewrite the equation as:
$4u^3 + 4u^2 - 2u^2 - 2u + 2u - 1 = 0$

Now, factor by grouping:
$(4u^3 + 4u^2) + (-2u^2 - 2u) + (2u - 1) = 0$

Factor out the common terms in each group:
$4u^2(u + 1) - 2u(u + 1) + (2u - 1) = 0$

Notice that $4u^2(u + 1) - 2u(u + 1)$ can be factored further:
$u(4u^2 - 2)(u + 1) + (2u - 1) = 0$

Factor out $(2u - 1)$ :
$(2u - 1)(2u^2(u + 1) + 1) = 0$

Now, we have two factors to consider: $(2u - 1)$ and $(2u^2(u + 1) + 1)$ . Set each factor equal to zero and solve for $u$ .

First, solve $2u - 1 = 0$ for $u$ :
$2u - 1 = 0 \Rightarrow u = \frac{1}{2}$

Now, solve $2u^2(u + 1) + 1 = 0$ for $u$ . This is a quadratic in terms of $u^2$ .
$2u^2(u + 1) + 1 = 0 \Rightarrow 2u^3 + 2u^2 + 1 = 0$

This equation does not factor nicely, and it does not have rational roots. We can use the cubic formula or numerical methods to find the roots, but they will not be real since the discriminant is negative. Thus, the only real solution for $u$ is $u = \frac{1}{2}$ .

Now, recall that $u = \cos(2x)$ . Substitute $u = \frac{1}{2}$ back into this equation to solve for $x$ .
$\cos(2x) = \frac{1}{2}$

Solve for $2x$ by finding the angles whose cosine is $\frac{1}{2}$ . These angles are $60^{\circ}$ and $300^{\circ}$ .
$2x = 60^{\circ} \quad \text{or} \quad 2x = 300^{\circ}$

Divide each angle by 2 to solve for $x$ .
$x = \frac{60^{\circ}}{2} = 30^{\circ} \quad \text{or} \quad x = \frac{300^{\circ}}{2} = 150^{\circ}$