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## Answers

__<__x < 360°Just as with linear equations, I'll first isolate the variable-containing term: sin(x) + 2 = 3

sin(x) = 1Now I'll use the reference angles I've memorized: x = 90°Solve tan2(x) + 3 = 0 for 0°

__<__x < 360°There's the temptation to quickly recall that the tangent of 60° involves the square root of 3 and slap down an answer, but this equation doesn't actually have a solution:tan2(x) = –3How can the square of a trig function evaluate to a negative number? It can't!no solutionSolve on 0°

__<__x < 360°To solve this, I need to do some simple factoring:Now that I've done the algebra, I can do the trig. From the first factor, I get x = 90° and x = 270°. From the second factor, I get x = 30° and x = 330°.x = 30°, 90°, 270°, 330° Copyright © Elizabeth Stapel 2010-2011 All Rights ReservedSolve sin2(x) – sin(x) = 2 on 0°

__<__x < 360°This is a quadratic in sine, so I can apply some of the same methods:sin2(x) – sin(x) – 2 = 0

(sin(x) – 2)(sin(x) + 1) = 0

sin(x) = 2 (not possible!) or sin(x) = –1Only one of the factor solutions is sensible. For sin(x) = –1, I get:x = 270°Solve cos2(x) + cos(x) = sin2(x) on 0°

__<__x < 360°I can use a trig identity to get a quadratic in cosine:cos2(x) + cos(x) = sin2(x)

cos2(x) + cos(x) = 1 – cos2(x)

2cos2(x) + cos(x) – 1 = 0

(2cos(x) – 1)(cos(x) + 1) = 0

cos(x) = 1/2 or cos(x) = –1The first trig equation, cos(x) = 1/2, gives me x = 60° and x = 300°. The second equation gives me x = 180°. So my complete solution is: ADVERTISEMENT x = 60°, 180°, 300°Solve sin(x) = sin(2x) on 0°

__<__x < 360°I can use a double-angle identity on the right-hand side, and rearrange and simplify; then I'll factor:sin(x) = 2sin(x)cos(x)

sin(x) – 2sin(x)cos(x) = 0

sin(x)(1 – 2cos(x)) = 0

sin(x) = 0 or cos(x) = 1/2I can The sine wave is zero at 0°,180°, and 360°. The cosine is 1/2 at60°, and thus also at 360° – 60° = 300°. So the complete solution is:x = 0°, 60°, 180°, 300°, 360°Solve sin(x) + cos(x) = 1 on 0°

__<__x < 360°Hmm... I'm really not seeing anything here. It sure would have been nice if one of these trig expressions were squared...Well, why don't I square both sides, then, and see what happens?(sin(x) + cos(x))2 = (1)2

sin2(x) + 2sin(x)cos(x) + cos2(x) = 1

[sin2(x) + cos2(x)] + 2sin(x)cos(x) = 1

1 + 2sin(x)cos(x) = 1

2sin(x)cos(x) = 0

sin(x)cos(x) = 0Huh; go figger: I squared, and got something that I could work with. Nice!From the last line above, either sine is zero or else cosine is zero, so my solution appears to be:x = 0°, 90°, 180°, 270°However (and this is important!), I squared to get this solution, so I need to check my answers in the original equation, to make sure that I didn't accidentally create solutions that don't actually count. Plugging back in, I see:sin(0°) + cos(0°) = 0 + 1 = 1 (this solution works)

sin(90°) + cos(90°) = 1 + 0 = 1 (this one works, too)

sin(180°) + cos(180°) = 0 + (–1) = –1 (oh;okay, so this one does NOT work)

sin(270°) + cos(270°) = (–1) + 0 = –1 (this one doesn't work, either)So the actual solution is:x = 0°, 90°Note that I could have used the double-angle identity for sine, in reverse, instead of dividing off the 2 in the next-to-last line in my computations. The answer would have been the same, but I would have needed to account for the solution interval:2sin(x)cos(x) = sin(2x) = 0Then 2x = 0°, 180°, 360°, 540°, etc, and dividing off the 2 from the x would give me x = 0°, 90°, 180°, 270°, which is the same almost-solution as before. After doing the necessary check (because of the squaring) and discarding the extraneous solutions, my final answer would have been the same as before.