Inverse trig. (1 Viewer)

[xwrathbringerx]

Could someone please show me how to solve these equations:

a) arccos(x) + arccos(x*sqrt(3)) = pi/2

b) 2arcsin(x*sqrt(6)) + arcsin(4x) = pi/2

 

[kaz1]

1.cos[arccos(x) + arccos(x*sqrt(3))]=cos(pi/2)
cos[arccos(x)]cos[arccos(x*sqrt(3))]-sin[arccos(x)]sin[arccos(x*sqrt(3))]=0
x.xrt3-rt(1-x²)rt(1-3x²=0
I reckon you can do the rest
Bolded bit: draw a right angled triangle where cos^-1f(x) is the angle.

2. Follows a similar method, you can choose to sin both sides or cos both sides.

 

[Sunyata]

Duuuude how do you solve 2. I get this HUGE equation! 7744x^10 - 3008x^6 + 176 x^5 + 240x^4 - 8x^2 - 8x + 1 = 0