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

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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#