# Elementary Equations – Problems

## 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.