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How do you solve #sqrt3 sinx-cosx=1#?

Answer 1

Victoria Adebayo

Divide both sides by 2, use the identity:

#sin(x - phi) = sin(x)cos(phi) - cos(x)sin(phi)#

and then solve for x.

Given: #sqrt(3)sin(x) - cos(x) = 1" [1]"#

Divide both sides of equation [1] by 2:

#sqrt(3)/2sin(x) - (1/2)cos(x) = 1/2" [2]"#

The left side of equation [2] is the same form as the right side of the following trigonometric identity:

#sin(x - phi) = sin(x)cos(phi) - cos(x)sin(phi)" [3]"#

Where #cos(phi) = sqrt(3)/2# and #sin(phi) = 1/2#; we know this angle to be #pi/6#. Therefore, we can substitute #sin(x - pi/6)# into the left side of equation [2]:

#sin(x - pi/6) = 1/2" [4]"#

Use the inverse sine function on both sides of equation [4]:

#x - pi/6 = sin^-1(1/2)" [5]"#

The sine function is #1/2# at #theta = pi/6 and (5pi)/6#

#x - pi/6 = { (pi/6 + 2npi), ((5pi)/6 + 2npi) :}#

#x = { (pi/3 + 2npi), (pi + 2npi) :}#

where n is any positive or negative integer including 0.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.