Solve 4cos^2x-3=0 for all real values of x

x=30°, 150°, 210° & 330°

Step-by-step explanation:

4cos²x - 3 = 0  

=> Add 3 to both sides, then  

4cos²x = 3  

=> Divide both sides by 4, then  

cos²x = 3/4  

=> Square root both sides, then  

cos x = ± √(3/4)  

cos x = ± √3/2  

cos x = - √3/2 or √3/2  

=> Starting cos x = - √3/2, we calculate thus:  

x = cos^-1 (- √3/2)  

x = - cos^-1 (√3/2)  

=> In trigonometry, cos^-1 of √3/2 is 30° but the -ve arc - cos of a value denotes that the real value of x here can only be obtained from the 2nd & 3rd Quadrant of the unit circle, where the values of cos are -ve, then  

2nd Quadrant: 180° - 30° = 150°...  

3rd Quadrant: 180° + 30° = 210°...  

=> Moving on to x = cos^-1 (√3/2), we calculate thus:  

x = 30°  

=> In the unit circle of the 4 Quadrants, the 4th Quadrant has the cos of all values to be +ve, then  

4th Quadrant: 360° - 30° = 330°...  

Hence the real values of x in degrees are 30°, 150°, 210° & 330°

the value of v.w= -12.

step-by-step explanation:

we are given three vector r,v and w as:

r = < 5, -5, -2> ; v = < 2, -8, -8> ; w = < -2, 6, -5>

we know that r,v and w could also be represented as:

r=5i-5j-2k

v=2i-8j-8k

and w= -2i+6j-5k

where i,j and k are the unit vectors of x,y and z-axis respectively.

also we know that it has the following property:

i.i=1 ; j.j=1; k.k=1

and i.=j.i=i.k=k.i=j.k=k.j=0

i.e. when we multiply two vectors or we find it's dot product we need to only multiply the components of x coordinates with each other; y with each other and z with each other.

hence, v.w= (2i-8j- -2i+6j-5k)

                  = 2×(-2)+(-8)×(6)+(-8)×(-5)

                  = -4-48+40

                  = -12

hence, v.w= -12

341

step-by-step explanation:

all you have to do is 9x31 (which equals 279) and then 20x31 (which equals 620). then you have to subtract. 620-279=341.