Homework Question: solve the following equation:
sin x cos x = sqrt3/4

Question

Homework Question: solve the following equation:
sin x cos x = sqrt3/4

mathmale · Answer

The first way in which i'd try solving this problem would be to  re-write the left side.  Look for a trig identity that involves sin x cos x.

ellecullen · Answer

I did this: sqrt3/4 = 1/2 x sqrt3/2
sqrt3/2= cos(pi/6) 
then sin(x) x sqrt3/2) = sqrt3/4
sin(x) = (sqrt3/4) / (sqrt3/2) 
sin(x) = 1/2

mathmale · Answer

Have you found the value of x?  
In either case, how many solutions could you expect?  1?  2?  3?  
Have you checked your answer(s)?

mathmale · Answer

I do agree that your sin x = 1/2 leads to a solution.  But what is that solution?  And, again, is this the only solution?

ellecullen · Answer

My solution is pi/6 according to the unit circle stuff. It looks as if there is one solution: pi/6

mathmale · Answer

Would you mind checking whether or not 5Pi/6 is a solution?

ellecullen · Answer

5pi/6 is a solution for sin(x) = 1/2 meanwhile its cosine is a -(sqrt3/2). In this case, the -(sqrt3/2) does not work.

mathmale · Answer

So, I hear you saying that Pi/6 is a solution and 5Pi/6 is not.  Is that what you meant?

ellecullen · Answer

yes

myininaya · Answer

Also sin(x)cos(x) can be written as 1/2 *2sin(x)cos(x)=1/2 * sin(2x)

hero · Answer

$$\sin(x)\cos(x)=\frac{\sqrt{3}}{4}$$


$$4\sin(x)\cos(x)=\sqrt{3}$$

$$(4\sin(x)\cos(x))^2=(\sqrt{3})^2$$

$$16\sin^2(x)\cos^2(x)=3$$

$$16\sin^2(x)(1−\sin^2(x))=3$$

$$16\sin^2(x)−16\sin^4(x)=3$$


Let $y = \sin^2(x)$

$16y−16y^2=3$


$−16y^2+16y−3=0$


$16y^2−16y+3=0$


$16y^2−(12+4)y+3=0$


$16y^2−12y−4y+3=0$


$4y(4y−3)−1(4y−3)=0$


$(4y−3)(4y−1)=0$


$4y−3=0$

$4y−1=0$


Replace y with $\sin^2(x)$

$4\sin^2(x)−3=0$


$$4\sin^2(x)−1=0$$

hero · Answer

$$\sin^2(x) = \frac{3}{4}$$
$$\sin^2(x) = \frac{1}{4}$$

$$\sin(x) = \pm \frac{\sqrt{3}}{4}$$
$$\sin(x) = \pm \frac{1}{2}$$

only positive solutions for x