Hello, I need someone to check it for me, if I have done everything correct?
$$5\cos ^{2}a-7cosa-6=0$$
Let's say that cosa=m
$$5m ^{2}-7m-6=$$
$$D=(-7)^{2}-4*5*(-6)=49+120=169$$
$$m _{1}=\frac{ 7+13 }{ 10 }=2$$
$$m _{2}=\frac{ 7-13 }{ 10 }=-0.6$$

$$cosa=2 $$ it is not the sollution and
$$cosa=-0.6 $$ So, the problem starts here. Is -0.6 an answer or do I need to write like this
cosa=-0.6
a=$$\alpha=\pm \arccos \alpha +2 \pi k$$

Question

Hello, I need someone to check it for me, if I have done everything correct?
$$5\cos ^{2}a-7cosa-6=0$$
Let's say that cosa=m
$$5m ^{2}-7m-6=$$
$$D=(-7)^{2}-4*5*(-6)=49+120=169$$
$$m _{1}=\frac{ 7+13 }{ 10 }=2$$
$$m _{2}=\frac{ 7-13 }{ 10 }=-0.6$$

$$cosa=2 $$ it is not the sollution and
$$cosa=-0.6 $$ So, the problem starts here. Is -0.6 an answer or do I need to write like this
cosa=-0.6
a=$$\alpha=\pm \arccos \alpha +2 \pi k$$

kamille · Answer

@stamp

sirm3d · Answer

i agree with your answer.

kamille · Answer

well, is -0,6 an answer?

anonymous · Answer

you would have to sub back in

anonymous · Answer

so where you got $$m_1,m_2$$
would be 
$$cos(\alpha_1),cos(\alpha_2)$$

anonymous · Answer

then find what alpha is by taking the arcs of each

hartnn · Answer

is the Question , give general values of 'a' , or just solve for 'a' or any interval is given ?
also, wouldn't it be 
$a= \arccos(-0.6)\pm 2\pi k$

kamille · Answer

No, there is no interval: just solve the equation

anonymous · Answer

it'd be a general solution of the values

sirm3d · Answer

$$\alpha=\pm arccos(-0.6) + 2k\pi$$

hartnn · Answer

so, you have to write an general expression, $a= \arccos(-0.6)\pm 2\pi k$

kamille · Answer

and the most confusing part of the problem is the second part:
"Find sin alpha, if we know that alpha belongs to (pi; 3pi/2). I know i can solve sin aplha from sin^2a+cos^2a=1, but it would be much more easier if I just leave a=-0.6 in the first part.

hartnn · Answer

why $\pm$ is with arccos and not with 2kpi ?

sirm3d · Answer

there are two sets of solutions, one $$\arccos (-0.6)+2k\pi$$ and the other $$-\arccos(-0.6)+2k\pi$$

sirm3d · Answer

@Kamille 's factorization is correct, @stamp

hartnn · Answer

but a=-0.6 is not correct solution for 1st part.

kamille · Answer

so, do I need to find value of alpha in interval of {pi;3pi/2)?

hartnn · Answer

(pi; 3pi/2) means 3rd quadrant, so when you find sin a from that identity, just add a negative sign because sin is negative in 3rd quadrant.

stamp · Answer

$$5x^2-7x-6$$$$(5x+3)(x-2)$$
$$5cosx+3=0$$$$x=cos^{-1}(-3/5)$$
$$cosx-2=0$$$$x=cos^{-1}(2)$$

How about that

sirm3d · Answer

the range of cosine is between -1 and +1, @stamp

sirm3d · Answer

|dw:1362905898862:dw|

sirm3d · Answer

that's why there are two values of alpha corresponding to arccosine -0.6.
$$a=\pm \arccos(-0.6) + 2k\pi\quad,\text{k an integer}$$

kamille · Answer

So i have:

a=$$a=\pm \arccos (-0.6) + 2 \pi k$$
Can someone remind me how to get radians of 0.6?

sirm3d · Answer

you have to use your calculator.

kamille · Answer

well, at school we had a formula (which I have forgotten) like if you have 1/2 you can divide(or multiply) something by pi etc, do you know it?

sirm3d · Answer

i think the formula works only for special angles.

hartnn · Answer

could you do 2nd part, finding sin a ? 
and numerical value of arccos(-0.6) ?