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How do you solve the following, please show the steps solve for x, 0 less than and/or...
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`tan(x+5) = -3`
`=>(x+5) = tan^(-1)(-3)`
`=>x+5 = -71. 6ᵒ or, (180-71.6)ᵒ and, (360-71.6)ᵒ`
[as tanθ is positive in 1st and 3rd quadrant; negative in 2nd or fourth quadrant]
`=>x+5 = 108.4ᵒ and 288.4ᵒ`
`x =(108.4-5)ᵒ and (288.4-5)ᵒ `
`= 103.4ᵒ degrees and 283.4ᵒ degrees`
Hence solution of the equation tan(x+5) = -3 [0<x<360] yields x == 103.4ᵒ and 283.4ᵒ
Posted by llltkl on May 20, 2013 at 1:47 AM (Answer #1)
Thanks for your response.
For every trigonometric function, there is an inverse function that works in the reverse. These inverse functions have the same name but with 'arc' in front. (On some calculators the arcsin button may be labelled asin, or sometimes `sin ^(-1)` ) So the inverse of sin is arcsin etc. When we see "arcsin A", we understand it as "the angle whose sin is A"
Thus, sin30 = 0.5 means: the sin of 30 degrees is 0.5; arcsin 0.5 = 30 means: the angle whose sin is 0.5 is equal to 30 degrees.
Tangent is defined as = sin/cos; its values are also periodic and the sign of the tangent can be estimated by:
–π to –π/2: sine is negative and cosine is negative, so tangent is positive
–π/2 to 0: sine is negative but cosine is positive, so tangent is negative
π to 3π/2: sine is negative and cosine is negative, so tangent is positive
3π/2 to 2π: sine is negative but cosine is positive, so tangent is negative
The angle 246.6 falls in the range π to 3π/2, hence its tangent must have been positive. That is not the case here. So that answer cannot be right.
Actually tangent is a periodic function of period length = π (180 degrees).
Posted by llltkl on May 20, 2013 at 6:33 PM (Answer #5)
The answer given by the teacher was 103.4 degrees and 246.6 degrees so the second part is not correct unless my teacher is wrong or perhaps 360 -71.6 is not right. I still do not understand how you know to work the first step of this problem. I am in middle school so can you please explain in more detail. My teacher gave us the answers but I need to figure out how to work the problem. Thank you!
Posted by kelcsl on May 20, 2013 at 2:25 AM (Answer #2)
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