Solve #11sin(theta)-2=2# for a value of #theta# when #0 <= theta <= pi/2#.?

Question

I think I know how to do the first one but I'm unsure about how to do the second problem

1685 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-10-26T22:47:10

#theta = 0.659#

#5cos2theta+7=cos2theta+8#

Simplifies to (simply by rearranging):
#4cos2theta=1#

We know #theta=1.32 # by using inverse cos, and now we need to find out #2theta#.

#2theta=1.32#
#theta = 0.659#

Answer 2 · 2017-10-26T22:49:47

Given: #5cos(2theta)+7=cos(2theta)+8; 0 <= theta <= pi/2#

Subtract 7 from both sides:

#5cos(2theta)=cos(2theta)+1; 0 <= theta <= pi/2#

Subtract #cos(2theta)# from both sides:

#4cos(2theta)=1; 0 <= theta <= pi/2#

Divide both sides by 4:

#cos(2theta)=1/4; 0 <= theta <= pi/2#

Use the inverse cosine on both sides:

#2theta= cos^-1(1/4); 0 <= theta <= pi/2#

Divide both sides by 2:

#theta = 1/2cos^-1(1/4); 0 <= theta <= pi/2#

#theta = 0.659#