1. Prove x is an integer and state it's value in: (3sinx + cosx)^2 + (sinx-3cosx)^2
2. Show 6sin^2x = 4 + cosx is the same as 6cos^2x + cosx - 2 = 0
3. Show 7sin^2θ +sinθcosθ = 6 is the same as tan^2θ + tanθ - 6 = 0
4. Prove: sinA/1+cosA + 1+cosA/sinA ≡ 2/sinA
5. Show that 3+sin^2θ / cosθ-2 = 3cosθ
6. Solve cos(3x+50) = (root(3))/2 between -180≤x≤360
7. Solve 6tanxsinx=5
8. Solve: 5cos^2x - 13sinx + 1 = 0 (hence solve 5cos^2(x+30) - 13sin(x+30) + 1 = 0) range= 0<x<360
Diapositiva 2
1. Prove x is an integer and state it's value in: (3sinx + cosx)^2 + (sinx-3cosx)^2
Expand: 9sin^2x + 6sinxcosx + cos^2x + sin^2x - 6sinxcosx + 9cos^2x
Collect together: 10sin^2x + 10cos^2x
Factor out 10: 10(sin^2x + cos^2x)
Use identity that sin^2x + cos^2x ≡ 1 : 10(1) = 10
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Diapositiva 3
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2. Show 6sin^2x = 4 + cosx is the same as 6cos^2x + cosx - 2 = 0
sin^2x + cos^2x≡ 1 , so sin^2x≡1-cos^2x.
Here we can say: 6(1-cos^2x) = 4+cosx
Expand and rearrange into quadratic: 6cos^2x + cosx - 2 = 0
Solve for x=1/2 or -2/3
Diapositiva 4
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3. Show 7sin^2θ +sinθcosθ = 6 is the same as tan^2θ + tanθ - 6 = 0
Start at answer: Use identity tanθ = sinθ/cosθ ; tan^2θ + tanθ - 6 = 0 is same as sin^2θ / cos^2θ + sinθ/cosθ - 6 = 0
Now remove denominators: sin^2θ + sinθcosθ - 6cosθ = 0
We use identity sin^2θ + cos^2θ ≡ 1 and rearrange to cos^2θ ≡ 1- sinθ and plug in : sin^2θ + sinθcosθ - 6(1-sin^2θ)
Same as: sin^2θ + sinθcosθ - 6 + 6sin^2θ = 0 , collect together for: 7sin^2θ+sinθcosθ-6=0
Same as: 7sin^2θ +sinθcosθ = 6
5. Show that 3+sin^2θ / cosθ-2 = 3cosθ
Remove denominator: 3+sin^2θ=3cosθ(cosθ-2)
Use identity sin^2θ + cos^2θ ≡ 1 and rearrange to sin^2θ ≡ 1- cosθ to say 3+(1-cos^2θ) = 3cosθ(cosθ-2)
Same as: 4-cos^2θ = 3cos^2θ - 6cosθ
Move to 1 side and collect: 4cos^2θ - 6cosθ - 4 = 0
Solve quadratic: cos=-1/2 or 2
Cos must be -1/2 as -1≤cos≤1
7. Solve 6tanxsinx=5
First show that 6tanxsinx=5 is the same as 6cos^2x+5cosx-6=0 : 6(sin/cos)sin=5 , 6sin^2x/cosx =5, 6sin^2x=5cosx, 6(1-cos^2x)=5cos, 6cos^2x+5cosx-6=0
Now solve: solve quadratic for cosx=2/3 or -3/2 , but the it must be 2/3 as -1≤cos≤1 and find all other values which right now i cba to do so sry future self
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Diapositiva 9
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5cos^2x - 13sinx + 1 = 0 (range= 0<x<360)
cos^2x + sin^2x ≡ 1
5(1 - sin^2x) - 13sinx + 1 = 0 , 5 - 5sin^2x - 13sin + 1 , 5sin^2x + 13sinx - 6 = 0
Solve 5sin^2x + 13sinx - 6 = 0 as a quadratic to get sin = 2/5 or -3.
Sin-1(2/5) = 23.6 , 180-23.6 = 156.4. +/- 360 but gives values not in range. Sin-1(-3) is invalid as sin is from -1 to 1.
Hence 5cos^2(x+30) - 13sin(x+30) + 1 = 0
x+30 = 23.6 , 156.4
x= -6.4 , 126.4 +/-360 = 353.6
Range= 0<x<360 , so answers = 126.4 , 353.6
