Elementary Equations – Problems
1.Find the values of θ which satisfy rsinθ = 3 and r = 4 (1 + sinθ), 0 ≤ θ ≤ 2π.
solution:
rsinθ = 3 and
r = 4 (1 + sinθ), 0 ≤ θ ≤ 2π
rsinθ = 3
r = 3/sinθ
3/sinθ = 4 (1 + sinθ)
3 = sinθ 4 (1 + sinθ)
3 =4sin²θ + 4sin θ
4sin²θ + 4sin θ – 3 = 0
4sin²θ + 6sin θ – 2sin θ – 3 = 0
2sinθ( 2sinθ + 3) – 1(2sinθ + 3) = 0
(2sinθ – 1) (2sinθ + 3) = 0
(2sinθ – 1) = 0
2sinθ = 1
Sinθ = ½
θ = π/6, 5π/6
2sinθ + 3 = 0
2sinθ = -3
Sinθ = -3/2 ( not passible)
2. Solve \({{16}^{{{\sin }^{2}}x}}+{{16}^{{{\cos }^{2}}x}}=10\), 0 ≤ θ ≤ 2π.
Solution:
\({{16}^{{{\sin }^{2}}x}}+{{16}^{{{\cos }^{2}}x}}=10\),
Let us consider \({{16}^{{{\sin }^{2}}x}}=t\),
\({{16}^{{{\sin }^{2}}x}}+{{16}^{1-{{\sin }^{2}}x}}=10\),
\({{16}^{{{\sin }^{2}}x}}+16\times {{16}^{-{{\sin }^{2}}x}}=10\),
\({{16}^{{{\sin }^{2}}x}}+16\times \frac{1}{{{16}^{{{\sin }^{2}}x}}}=10\),
\(t+16\times \frac{1}{t}=10\),
\({{t}^{2}}-10t+16=0\),
\({{t}^{2}}-8t-2t+16=0\),
\(t\left( t-8 \right)-2\left( t-8 \right)=0\),
\(\left( t-8 \right)\left( t-2 \right)=0\),
\(t=8,2\).
Case(i): Put t = 8
\({{16}^{{{\sin }^{2}}x}}=t\),
\({{16}^{{{\sin }^{2}}x}}=8\),
\({{2}^{4}}^{{{\sin }^{2}}x}={{2}^{3}}\),
Bases are equal then power should be equal
\(4{{\sin }^{2}}x=3\),
\({{\sin }^{2}}x=\frac{3}{4}\),
\(\sin x=\sqrt{\frac{3}{4}}\),
\(\sin x=\pm \frac{\sqrt{3}}{2}\),
\(\sin x=\frac{\sqrt{3}}{2},\ then\ x=\frac{\pi }{3},\frac{2\pi }{3}\),
\(\sin x=-\frac{\sqrt{3}}{2},\ then\ x=\frac{4\pi }{3},\frac{5\pi }{3}\).
Case(ii): Put t = 2
\({{16}^{{{\sin }^{2}}x}}=t\),
\({{2}^{4}}^{{{\sin }^{2}}x}=2\),
Bases are equal then power should be equal
\(4{{\sin }^{2}}x=1\),
\({{\sin }^{2}}x=\frac{1}{4}\),
\(\sin x=\pm \frac{1}{2}\),
\(\sin x=\frac{1}{2},\ then\ x=\frac{\pi }{6},\frac{5\pi }{6}\),
\(\sin x=-\frac{1}{2},\ then\ x=\frac{7\pi }{6},\frac{11\pi }{6}\),
Hence, there will be eight solutions in all.