This is grade 12 trig identities

## Your answer

 Your name to display (optional): Email me at this address if my answer is selected or commented on: Privacy: Your email address will only be used for sending these notifications. Anti-spam verification: To avoid this verification in future, please log in or register.

## 1 Answer

The solution is x=arcsin(0.25), so we need to find an equation with the solution 3arcsin(0.25).

So x/3=arcsin(0.25) and sin(x/3)=0.25.

sin(x)=sin(2x/3+x/3)=sin(2x/3)cos(x/3)+cos(2x/3)sin(x/3)=0.25.

2sin(x/3)cos²(x/3)+(1-2sin²(x/3))sin(x/3)=0.25.

2sin(x/3)(1-sin²(x/3))+sin(x/3)-2sin³(x/3)=0.25.

3sin(x/3)-4sin³(x/3)=0.25. This cubic equation needs to be solved for x/3.

It can be written 16sin³(x/3)-12sin(x/3)+1=0.

If y=x/3, 16sin³y-12siny+1=0 is an equation with the required solution given by y.

x=arcsin(0.25)=0.25268 or 2.88891 approx, so y=x/3=0.084227 or 0.96297, siny=0.0841272 or 0.820892 which is a solution of the cubic. The third solution to the cubic is siny=-0.905019, y=5.151761, if the domain applies to y.

by Top Rated User (642k points)

1 answer
0 answers
1 answer
1 answer
1 answer
1 answer
0 answers
1 answer
1 answer
1 answer