Topic: Trig Equations Homework Help?**Question:**
Solve algebraically for 0 ≤ x ≤ 360:
a) 3sin2x° + 1 = 2
b) 2sin 3x° = 1
c) sin^2x° = 3 / 4
Can someone please help me with these 3 questions as I have a test coming up, thanks.

June 27, 2019 / By Abaigeal

Fundamental formula for sin: sinx = siny x = y + k360 x = (180 - y) + k360 a) 3sin2x = 1 sin2x = 1/3 2x = sin(-1)(1/3) + k360 2x = 19 + k360 x = (19 + k360)/2 k = 0, x = 19/2 = 9.5 k = 1, x = 379/2 = 189.5 k = 2, x = 739/2 = 369.5 --> this is over 360°, the question specifies 0 ≤ x ≤ 360 so you cannot take this answer, it is wrong or 2x = 161 + k360 x = (161 + k360)/2 k = 0, x = 161/2 = 80.5 k = 1, x = 521/2 = 260.5 x = 9.5°, 189.5°, 80.5°, 260.5° b) 2sin3x = 1 sin3x = 1/2 3x = sin^-1(1/2) 3x = 30 + k360 x = (30+k360)/3 k = 0, x = 30/3 = 10 k = 1, x = 390/3 = 130 k = 2, x = 750/3 = 250 3x = 150 + k360 x = (150 + k360)/3 k = 0, x = 150/3 = 50 k = 1, x = 510/3 = 170 k = 2, x = 870/3 = 290 x = 10°, 130°, 250°, 50°, 170°, 290° c) sin2x = 3/4 2x = sin^-1(3/4)+k360 2x = 49 + k360 x = (49 + k360)/2 k = 0, x = 49/2 = 24.5 k = 1, x = 409/2 = 204.5 2x = 131 + k360 x = (131 + k360)/2 k = 0, x = 131/2 = 65.5 k = 1, x = 491/2 = 245.5 x = 24.5°, 204.5°, 65.5°, 245.5°

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Did you like the answer? Recall that: sinθ = y/r cosθ = x/r tanθ = y/x You can derive these from reciprocal identities: cscθ = r/y secθ = r/x cotθ = x/y Since the point is P(3, 6), then, by the Pythagorean Theorem, we get: r = √(3² + 6²) = √45 So we have that: x = 3, y = 6, r = √45 Then, by the definition of the trig functions, we get: sinθ = y/r = 6/√45 = (18√5)/45 = (2√5)/5 cosθ = x/r = 3/√45 = (9√5)/45 = √5/5 tanθ = y/x = 6/3 = 2 cscθ = r/y = √45/6 = (3√5)/6 = √5/2 secθ = r/x = √45/3 = (3√5)/3 = √5 cotθ = x/y = 3/6 = 1/2 I hope this helps!

remember that x^2 + y^2 = r^2, in standard position so 3^2 + 6^2 = 9 + 36 = 45 r^2 = 45 ==> r = sqrt(45) = 3 sqrt(5) now use the definitions of the functions: sin t = y / r = 6 / (3sqrt(5) = 2 / sqrt(5) = 2sqrt(5) / 5 cos t = x / r = 3 / (3sqrt(5)) = 1 / srqt(5) = sqrt(5) / 5 tan t = y / x = 6 / 3 = 2 csc t = r / y = 3sqrt(5) / 6 = sqrt(5) / 2 sec t = r / x = 3srqt(5) / 3 = sqrt(5) cot t = x / y = 3 / 6 = 1/2

a) 3sin(2x*)+1=2=> 3sin(2x*)=1=> sin(2x*)=1/3=> 2x=19.47122063..*=> x=9.735610317...*=> x=9* 44' 8" or 180*-9* 44' 8"=> x=9* 44' 8 " or 170* 15' 51" approximately. b), c) can be solved in the similar ways.

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sin is unfavourable in quadrants III and IV you ought to renowned the era the respond is being requested for (often the two [0 , 360) or all recommendations (the widely used sort of the answer, which incorporates all recommendations in the two Q3 and this fall)) one way: use the reference perspective, sin^-a million (3/4) -- in different words, use the valuable value of the trig function which will continually be in Q1 then, whilst you're searching for Q2, this is one hundred eighty - RA in Q3, this is one hundred eighty + RA for the time of this fall, this could be 360 - RA

👍 24 | 👎 -7

1. Change sin2x into 2sinxcosx using the double angle identity. 2sinxcosx = cosx 2sinxcosx - cosx = 0 common factor.... cosx(2 sinx - 1) = 0 so, either cosx = 0 or 2sinx - 1=0 If cosx=0 then you know x = pi/2 or 3pi/2 [(3/2)pi] if 2sinx -1 = 0 then sinx = 1/2 so x=pi/6 or 5pi/6 [(5/6)pi] <-- second answer comes from the CAST rule 2. yup, change cosx^2 into 1 - (sinx)^2 and then you'll have: 1 - (sinx)^2 - 3sinx - 3 = 0 -(sinx)^2 - 3sinx -2 = 0 get rid of the negative by multiplying through by -1 (sinx)^2 + 3sinx + 2 = 0 Now just think of sinx=a or some other variable: a^2 + 3a + 2 = 0 and you can factor this trinomial using two numbers that add to +3 and multiply to +2 (a + 2) (a + 1) = 0 so a = -2 or a = -1 But, since a really = sinx, that means sinx=-2 or sinx = -1 the first is impossible, and for the second you can just use the "shift/2nd function" sin on the calculator (or picture the graph of the sine curve) to get x = (3/2)pi

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