If cos2 x + 2cosx - 2 = 0, then cosx = _____.
cos²x+2cosx-2=0 , like this ?
yes
oh my bad
in this case its still a quadratic though.
Okay, so \(\normalsize\color{blue}{ \cos^2 x + 2\cos x - 2 = 0}\) :) For better vision let cos(x)=a, you come out with, \(\normalsize\color{blue}{ a^2 + 2a - 2 = 0}\) lets complete the square. \(\normalsize\color{blue}{ a^2 + 2a - 2 = 0}\) \(\normalsize\color{blue}{ a^2 + 2a +1 = 3}\) \(\normalsize\color{blue}{ (a+1)^2= 3}\) \(\normalsize\color{blue}{ a= -1±\sqrt{3}}\)
oh!!! thank you!!
yes phys, it is just a quadratic, technically, besides that for 'a' we can have 1 or 2 solutions, and for cosine function there are infinite number of solutions in many different intervals.
You welcome, @garcia22and
could you help me with another problem? please
yes, perhaps.
In triangle ABC, a = 12, angleB = 25°, and angleC = 45°. Find b.
sin(110)/12 = sin(25)/b
I mean A) 5.4 B) 6.6 C) 7.2 D) 22
No phys, angle A is not given yet, although, \(\normalsize\color{blue}{ ∠~A~~=~~180-(~~~∠~B + ∠~C~~)}\) .
You can't use the sine law yet, but after you find angle a, please go for it.
Sin(A)/a = Sin(B)/b = Sin(C)/c
so you solve for angle A. 180-25-45 = ?
\[\frac{ \sin110 }{ 12 }= \frac{ sinB }{ b }\]
right and we have Sin B
yes, we have the angle B, it is 25
when I get here what do I do next\[bsin110=5.071\]
to make it simpler you could just say 12sin(25)/sin(110) = B
I divide sin110?
think about it algebraicly
we want to isolate B
ok I got it, thanks!
Answer is 5.4
yep
cool!! another problem: Which of the following functions has a period of 4π ? A) y = sinx B) y = sinx C) y = sin2x D) y = sin4x I thought it was B
so for sin functions the period = 2pi/B
uh for answers a and b are the same?
oh sorry A is y=sin1/4x and B is sin1/2x
2pi/B = 4pi (period) solve for B get 1/2, so the answer is sin1/2x
ah okay so I was right, cool!
The vertical component of a force of 100 pounds that makes an angle of 30° with the horizontal is 50 pounds. Is this true?
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