i have two questions for you to check please!
6cos^2x+5cosx-4=0 where 0degrees < or equal to 0 which is < or equal to 360.
I got that this factors to (3x+4)(2x-1) so the answer is x= -4/3 and x= 1/2 ?? right?
and the second question:
determine sin2theta, cos2theta, and tan2theta. the given is sin x=3/5 and that cosx < 0. (this is where i get confused. how do i know what quadrant its in if cosx<0?)
If i put it in first i get 5 as hypotnuse 4 as adjacent side and 3 as opp? is that going in the right direction? please help
for the first you are not done
You are on the right track to treat it as a quadratic.
your factors are (3cosx +4)(2cosx - 1)=0
so cosx =-4/3 or cosx = 1/2
the first of these is not possible since the cosine lies between -1 and +1.
so cosx = 1/2
the cosine is positive in quadrants 1 and 4
so if cosx=1/2 x = 60º or x = 300º
for you second question....
<<given is sin x=3/5 and that cosx < 0>>
If you know the CAST rule, then the angle must be in quadrant #2 (between 90 and 180)
if cosx=3/5, you were right to draw a triangle and conclude that sinx must be 4/5
Now use the special angle relationships
sin 2x = 2sinxcosx = 2(4/5)(3/5) = ...
cos 2x = cos^2 x - sin^2 x = 9/25 - 16/25 = ...
and of course tan 2x = sin 2x/cos 2x so....
(pi/3 and 5pi/3 in radians)
That should have been
sin x = 3/5 and cosx = -4/5
I am sure you can make the necessary changes in my substitution
Also the "(pi/3 and 5pi/3 in radians)"
should have been at the end of the first solution.
For the first question:
The given equation is 6cos^2x+5cosx-4=0. To solve this equation, you correctly recognized that it can be factored as (3cosx+4)(2cosx-1)=0.
To find the values of x that satisfy this equation, you set each factor equal to zero:
3cosx+4=0 or 2cosx-1=0
Solving the first equation, you get:
3cosx=-4
cosx=-4/3
However, cosine values can only range between -1 and +1, so cosx=-4/3 is not a valid solution.
Next, you solve the second equation:
2cosx=1
cosx=1/2
This is a valid solution. Since the cosine is positive in quadrants 1 and 4, the values of x that satisfy cosx=1/2 are 60 degrees (quadrant 1) and 300 degrees (quadrant 4).
Therefore, the solutions to the equation are x=60 degrees and x=300 degrees.
For the second question:
You are given sinx=3/5 and cosx<0. Since the cosine is negative, the angle must lie in either quadrant 2 (between 90 and 180 degrees) or quadrant 3 (between 180 and 270 degrees).
To determine the quadrant, you can use the CAST rule. In this case, cosx<0 means the angle is in quadrant 2.
You drew a triangle with a hypotenuse of 5, adjacent side of 4 (since cosx is negative), and opposite side of 3. This confirms that sinx=3/5.
To find sin2x, cos2x, and tan2x, you can use the double angle formulas:
sin2x = 2sinxcosx = 2(3/5)(-4/5) = -24/25
cos2x = cos^2x - sin^2x = (-4/5)^2 - (3/5)^2 = 16/25 - 9/25 = 7/25
tan2x = sin2x/cos2x = (-24/25)/(7/25) = -24/7
So, sin2x = -24/25, cos2x = 7/25, and tan2x = -24/7.
Please note that there was a correction made regarding the values of sinx and cosx in the initial response.