sin[arcsin(1/5)+ arccos x] =1 find the value of x?

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This question is best solved in smaller parts. For instance,

sin (theta) = 1

theta = pi/2 or 90 degrees. I'm going to assume all answers for angles will be in radians.

Then, arcsin (1/5) + arccos x equals pi/2, since sin (pi/2)=1

arccos x = pi/2 - arcsin (1/5)

x = cos (pi/2 - arcsin (1/5))

Using the identity: cos (a - b) = cos a cos b + sin a sin b

x = cos (pi/2) cos (arcsin (1/5)) + sin (pi/2) sin (arcsin (1/5))

x = 0 * cos (arcsin (1/5)) + 1 * 1/5

**x = 1/5**

To solve this problem, you use that sin (pi/2) = 1

Therefore the entire expression on the inside of the sine brackets is equal to pi/2.

Then solve for x by taking the cos of both sides.

Use the difference identity for cosine to evaluate x.

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