Mathematics, 31.12.2019 01:31 jesus3426

Find the sum of the three smallest positive values of θ such that $4\cos^2(2\theta-\pi) =3$ .
(give your answer in radians.)

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Answer from: waltersamy30

∅= 13π/12

Step-by-step explanation:

Solve until you get cos(___) on one side and a value on the other side.

I got cos(2∅-π) = ±√(3)/2

I know that the only cos() value that gives me ±√(3)/2 are -5π/6, -π/6, and π/6 from looking at the unit circle; those 3 are the smallest value that would give me ±√(3)/2.

So i solve for what value of 2∅-π would give me -5π/6, -π/6, and π/6

by solving for

2∅₁-π = -5π/6, 2∅₂-π = -π/6, and 2∅₃-π = π/6

I got ∅₁=π/12, ∅₂=5π/12, ∅₃=7π/12.

The question wanted the sum of the three smallest value so i added it up and got 13π/12.

And thanks for peer review. Good catch on my mistakes.

Find the sum of the three smallest positive values of θ such that <img src=

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Answer from: breniljakenotro

$\frac{13}{12}\pi$

Step-by-step explanation:

$4\cos^2(2\theta-\pi)=3$

We want to first isolate the trig expression.

We will do this by first dividing both sides by 4 and then taking the square root of both sides.

$\cos^2(2\theta-\pi)=\frac{3}{4}$

$\cos(2\theta-\pi)=\pm \sqrt{\frac{3}{4}}$

Let's clean up the right hand side a little:

$\cos(2\theta-\pi)=\pm \frac{\sqrt{3}}{2}$

Break time to think about something to help us solve the above:

So $\cos(u)=\frac{\sqrt{3}}{2}$ when $u=\pm \frac{\pi}{6}$ in the one cycle of $\cos(u)$ .

So $\cos(u)=-\frac{\sqrt{3}}{2}$ when $u=\pm \frac{5\pi}{6}$ in the one cycle of $\cos(u)$ .

Back to it:

$\cos(2\theta-\pi)=\pm \frac{\sqrt{3}}{2}$

Solving one of those equations:

$\cos(2\theta-\pi)=\frac{\sqrt{3}}{2}$ when

$2\theta-\pi=\pm \frac{\pi}{6}+2\pi k$

Let's solve for $\theta$ .

Add $\pi$ on both sides:

$2\theta=\pm \frac{\pi}{6}+2\pi k+\pi$

Divide both sides by 2:

$\theta=\pm \frac{\pi}{12}+\pi k+\frac{\pi}{2}$

$\theta=\pm \frac{\pi}{12}+\frac{\pi}{2}+\pi k$

Now we also have to solve the other:

$\cos(2\theta-\pi)=-\frac{\sqrt{3}}{2}$ when

$2\theta-\pi=\pm \frac{5\pi}{6}+2\pi k$

Let's solve for $\theta$ .

Add $\pi$ on both sides:

$2\theta=\pm \frac{5\pi}{6}+2\pi k+\pi$

Divide both sides by 2:

$\theta=\pm \frac{5\pi}{12}+\pi k+\frac{\pi}{2}$

$\theta=\pm \frac{5\pi}{12}+\frac{\pi}{2}+\pi k$

So the full set of solutions is:

$\theta=\pm \frac{\pi}{12}+\frac{\pi}{2}+\pi k$

$\theta=\pm \frac{5\pi}{12}+\frac{\pi}{2}+\pi k$

---------------------------------------------------------------------------------

I'm going to evaluate these expressions for some integer :

k=-2 we have:

$\theta=\pm \frac{\pi}{12}+\frac{\pi}{2}+\pi (-2)$

$\theta=\frac{-17}{12}\pi \text{ or } \frac{-19}{12}\pi$

$\theta=\pm \frac{5\pi}{12}+\frac{\pi}{2}+\pi (-2)$

$\theta=\frac{-13}{12}\pi \text{ or } \frac{-23}{12}\pi$

k=-1 we have:

$\theta=\pm \frac{\pi}{12}+\frac{\pi}{2}+\pi (-1)$

$\theta=\frac{-5}{12}\pi \text{ or } \frac{-7}{12}\pi$

$\theta=\pm \frac{5\pi}{12}+\frac{\pi}{2}+\pi (-1)$

$\theta=\frac{-1}{12}\pi \text{ or } \frac{-11}{12}\pi$

k=0 we have:

$\theta=\pm \frac{\pi}{12}+\frac{\pi}{2}+\pi (0)$

$\theta=\frac{7}{12}\pi \text{ or } \frac{5}{12}\pi$

$\theta=\pm \frac{5\pi}{12}+\frac{\pi}{2}+\pi (0)$

$\theta=\frac{11}{12}\pi \text{ or } \frac{1}{12}\pi$

k=1 we have:

$\theta=\pm \frac{\pi}{12}+\frac{\pi}{2}+\pi (1)$

$\theta=\frac{19}{12}\pi \text{ or } \frac{17}{12}\pi$

$\theta=\pm \frac{5\pi}{12}+\frac{\pi}{2}+\pi (1)$

$\theta=\frac{23}{12}\pi \text{ or } \frac{13}{12}\pi$

Anyways the values will keep getting larger from here.

We can see that the smallest positive values happen when k=0 .

So we have the first smallest is: \frac{1}{12}\pi[/tex].

The second smallest is $\frac{5}{12}\pi$ .

The third smallest is $\frac{7}{12}\pi$ .

So the sum of these numbers are:

$\frac{1}{12}\pi+\frac{5}{12}\pi+\frac{7}{12}\pi$

$\frac{1+5+7}{12}\pi$

$\frac{13}{12}\pi$

Answer from: Quest

The amount used= 13.04- 1.9 = 11.14

Answer from: Quest

i think it might be a but you might want a second opinion

Alexa and bart are arguing about who will do the dishes. how can they make a fair decision selection

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