Exercise 5 : 1) Solve the following equations correct to one decimal place where necessary: $$ \begin{array}{ll} ext { a) } \sin heta=0,35 & ext { c) } \cos \left(2 heta+10^{\circ} ight)=0,866 \ ext { b) } an 2 heta=5 & ext { d) } \cos \left(x+20^{\circ} ight)=\sin 70^{\circ} \ ext { 2) Solve the following equations correct to one decimal place where necessary: } \ \begin{array}{ll} ext { a) } \sin ^{2} x-\sin x \cos x=0 & ext { g) } \cos ^{2} heta-\sin heta-1=0 \ ext { b) } 2 \sin x an x- an x=0 & ext { h) } 3 \sin ^{2} heta-\sin heta \cos heta-1=0 \ ext { c) } 8 \cos 2 heta=\frac{2}{\cos 2 heta} & ext { i) } \sin ^{2} heta-7 \sin heta \cos heta+2=0 \ ext { d) } \sin x \cos x-3 \cos x=0 & ext { j) }-\cos heta=\sin \left( heta-10^{\circ} ight) \ ext { e) } \sin \left( heta-20^{\circ} ight)=\cos heta & ext { k) } \cos 2 x=-\cos x\end{array}\end{array} $$

Question

Exercise 5 : 1) Solve the following equations correct to one decimal place where necessary: $$ \begin{array}{ll}	ext { a) } \sin 	heta=0,35 & 	ext { c) } \cos \left(2 	heta+10^{\circ}ight)=0,866 \ 	ext { b) } 	an 2 	heta=5 & 	ext { d) } \cos \left(x+20^{\circ}ight)=\sin 70^{\circ} \ 	ext { 2) Solve the following equations correct to one decimal place where necessary: } \ \begin{array}{ll}	ext { a) } \sin ^{2} x-\sin x \cos x=0 & 	ext { g) } \cos ^{2} 	heta-\sin 	heta-1=0 \ 	ext { b) } 2 \sin x 	an x-	an x=0 & 	ext { h) } 3 \sin ^{2} 	heta-\sin 	heta \cos 	heta-1=0 \ 	ext { c) } 8 \cos 2 	heta=\frac{2}{\cos 2 	heta} & 	ext { i) } \sin ^{2} 	heta-7 \sin 	heta \cos 	heta+2=0 \ 	ext { d) } \sin x \cos x-3 \cos x=0 & 	ext { j) }-\cos 	heta=\sin \left(	heta-10^{\circ}ight) \ 	ext { e) } \sin \left(	heta-20^{\circ}ight)=\cos 	heta & 	ext { k) } \cos 2 x=-\cos x\end{array}\end{array} $$

UpStudy AI · Accepted Answer

Solve the equation by following steps:
- step0: Solve for $$\theta$$:
  $$\theta +20=70$$
- step1: Move the constant to the right side:
  $$\theta =70-20$$
- step2: Subtract the numbers:
  $$\theta =50$$
Solve the equation $$ \tan(2\theta) = 5 $$.
Solve the equation by following steps:
- step0: Solve for $$\theta$$:
  $$\tan\left(2\theta \right)=5$$
- step1: Find the domain:
  $$\tan\left(2\theta \right)=5,\theta \neq \frac{\pi }{4}+\frac{k\pi }{2},k \in \mathbb{Z}$$
- step2: Use the inverse trigonometric function:
  $$2\theta =\arctan\left(5\right)$$
- step3: Add the period:
  $$2\theta =\arctan\left(5\right)+k\pi ,k \in \mathbb{Z}$$
- step4: Solve the equation:
  $$\theta =\frac{\arctan\left(5\right)}{2}+\frac{k\pi }{2},k \in \mathbb{Z}$$
- step5: Check if the solution is in the defined range:
  $$\theta =\frac{\arctan\left(5\right)}{2}+\frac{k\pi }{2},k \in \mathbb{Z},\theta \neq \frac{\pi }{4}+\frac{k\pi }{2},k \in \mathbb{Z}$$
- step6: Find the intersection:
  $$\theta =\frac{\arctan\left(5\right)}{2}+\frac{k\pi }{2},k \in \mathbb{Z}$$
Solve the equation $$ \tan(2\theta) = 5 $$.
Solve the equation by following steps:
- step0: Solve for $$\theta$$:
  $$\tan\left(2\theta \right)=5$$
- step1: Find the domain:
  $$\tan\left(2\theta \right)=5,\theta \neq \frac{\pi }{4}+\frac{k\pi }{2},k \in \mathbb{Z}$$
- step2: Use the inverse trigonometric function:
  $$2\theta =\arctan\left(5\right)$$
- step3: Add the period:
  $$2\theta =\arctan\left(5\right)+k\pi ,k \in \mathbb{Z}$$
- step4: Solve the equation:
  $$\theta =\frac{\arctan\left(5\right)}{2}+\frac{k\pi }{2},k \in \mathbb{Z}$$
- step5: Check if the solution is in the defined range:
  $$\theta =\frac{\arctan\left(5\right)}{2}+\frac{k\pi }{2},k \in \mathbb{Z},\theta \neq \frac{\pi }{4}+\frac{k\pi }{2},k \in \mathbb{Z}$$
- step6: Find the intersection:
  $$\theta =\frac{\arctan\left(5\right)}{2}+\frac{k\pi }{2},k \in \mathbb{Z}$$
Solve the equation $$ \sin(\theta - 20^{\text{o}}) = \cos(\theta) $$.
Solve the equation by following steps:
- step0: Solve for $$\theta$$:
  $$\sin\left(\theta -20\right)=\cos\left(\theta \right)$$
- step1: Move the expression to the left side:
  $$\sin\left(\theta -20\right)-\cos\left(\theta \right)=0$$
- step2: Transform the expression:
  $$-2\sin\left(\frac{\pi +40}{4}\right)\sin\left(\frac{\pi +40-4\theta }{4}\right)=0$$
- step3: Multiply both sides:
  $$-2\sin\left(\frac{\pi +40}{4}\right)\sin\left(\frac{\pi +40-4\theta }{4}\right)\left(-\frac{1}{2\sin\left(\frac{\pi +40}{4}\right)}\right)=0\times \left(-\frac{1}{2\sin\left(\frac{\pi +40}{4}\right)}\right)$$
- step4: Calculate:
  $$\sin\left(\frac{\pi +40-4\theta }{4}\right)=0\times \left(-\frac{1}{2\sin\left(\frac{\pi +40}{4}\right)}\right)$$
- step5: Calculate:
  $$\sin\left(\frac{\pi +40-4\theta }{4}\right)=0$$
- step6: Use the inverse trigonometric function:
  $$\frac{\pi +40-4\theta }{4}=\arcsin\left(0\right)$$
- step7: Calculate:
  $$\frac{\pi +40-4\theta }{4}=0$$
- step8: Add the period:
  $$\frac{\pi +40-4\theta }{4}=k\pi ,k \in \mathbb{Z}$$
- step9: Solve the equation:
  $$\theta =\frac{\pi +40}{4}+k\pi ,k \in \mathbb{Z}$$
Solve the equation $$ \theta = \arcsin(0.35) $$.
Solve the equation by following steps:
- step0: Solve for $$\theta$$:
  $$\theta =\arcsin\left(0.35\right)$$
Solve the equation $$ \sin^{2}(x) - \sin(x) \cos(x) = 0 $$.
Solve the equation by following steps:
- step0: Solve for $$x$$:
  $$\sin^{2}\left(x\right)-\sin\left(x\right)\cos\left(x\right)=0$$
- step1: Factor the expression:
  $$\sin\left(x\right)\left(\sin\left(x\right)-\cos\left(x\right)\right)=0$$
- step2: Separate into possible cases:
  $$\begin{align}&\sin\left(x\right)=0\&\sin\left(x\right)-\cos\left(x\right)=0\end{align}$$
- step3: Solve the equation:
  $$\begin{align}&x=k\pi ,k \in \mathbb{Z}\&x=\frac{\pi }{4}+k\pi ,k \in \mathbb{Z}\end{align}$$
- step4: Find the union:
  $$x=\left\{ \begin{array}{l}k\pi \\frac{\pi }{4}+k\pi \end{array}\right.,k \in \mathbb{Z}$$
Solve the equation $$ 2 \sin(x) \tan(x) - \tan(x) = 0 $$.
Solve the equation by following steps:
- step0: Solve for $$x$$:
  $$2\sin\left(x\right)\tan\left(x\right)-\tan\left(x\right)=0$$
- step1: Find the domain:
  $$2\sin\left(x\right)\tan\left(x\right)-\tan\left(x\right)=0,x\neq \frac{\pi }{2}+k\pi ,k \in \mathbb{Z}$$
- step2: Factor the expression:
  $$\tan\left(x\right)\left(2\sin\left(x\right)-1\right)=0$$
- step3: Separate into possible cases:
  $$\begin{align}&\tan\left(x\right)=0\&2\sin\left(x\right)-1=0\end{align}$$
- step4: Solve the equation:
  $$\begin{align}&x=k\pi ,k \in \mathbb{Z}\&x=\left\{ \begin{array}{l}\frac{\pi }{6}+2k\pi \\frac{5\pi }{6}+2k\pi \end{array}\right.,k \in \mathbb{Z}\end{align}$$
- step5: Find the union:
  $$x=\left\{ \begin{array}{l}k\pi \\frac{\pi }{6}+2k\pi \\frac{5\pi }{6}+2k\pi \end{array}\right.,k \in \mathbb{Z}$$
- step6: Check if the solution is in the defined range:
  $$x=\left\{ \begin{array}{l}k\pi \\frac{\pi }{6}+2k\pi \\frac{5\pi }{6}+2k\pi \end{array}\right.,k \in \mathbb{Z},x\neq \frac{\pi }{2}+k\pi ,k \in \mathbb{Z}$$
- step7: Find the intersection:
  $$x=\left\{ \begin{array}{l}k\pi \\frac{\pi }{6}+2k\pi \\frac{5\pi }{6}+2k\pi \end{array}\right.,k \in \mathbb{Z}$$
Solve the equation $$ \sin(x) \cos(x) - 3 \cos(x) = 0 $$.
Solve the equation by following steps:
- step0: Solve for $$x$$:
  $$\sin\left(x\right)\cos\left(x\right)-3\cos\left(x\right)=0$$
- step1: Factor the expression:
  $$\cos\left(x\right)\left(\sin\left(x\right)-3\right)=0$$
- step2: Separate into possible cases:
  $$\begin{align}&\cos\left(x\right)=0\&\sin\left(x\right)-3=0\end{align}$$
- step3: Solve the equation:
  $$\begin{align}&x=\frac{\pi }{2}+k\pi ,k \in \mathbb{Z}\&x \notin \mathbb{R}\end{align}$$
- step4: Find the union:
  $$x=\frac{\pi }{2}+k\pi ,k \in \mathbb{Z}$$
Solve the equation $$ 8 \cos(2\theta) = \frac{2}{\cos(2\theta)} $$.
Solve the equation by following steps:
- step0: Solve for $$\theta$$:
  $$8\cos\left(2\theta \right)=\frac{2}{\cos\left(2\theta \right)}$$
- step1: Find the domain:
  $$8\cos\left(2\theta \right)=\frac{2}{\cos\left(2\theta \right)},\theta \neq \frac{\pi }{4}+\frac{k\pi }{2},k \in \mathbb{Z}$$
- step2: Cross multiply:
  $$8\cos\left(2\theta \right)\cos\left(2\theta \right)=2$$
- step3: Simplify the equation:
  $$8\cos^{2}\left(2\theta \right)=2$$
- step4: Divide both sides:
  $$\frac{8\cos^{2}\left(2\theta \right)}{8}=\frac{2}{8}$$
- step5: Divide the numbers:
  $$\cos^{2}\left(2\theta \right)=\frac{1}{4}$$
- step6: Simplify the expression:
  $$\cos\left(2\theta \right)=\pm \sqrt{\frac{1}{4}}$$
- step7: Simplify the expression:
  $$\cos\left(2\theta \right)=\pm \frac{1}{2}$$
- step8: Separate into possible cases:
  $$\begin{align}&\cos\left(2\theta \right)=\frac{1}{2}\&\cos\left(2\theta \right)=-\frac{1}{2}\end{align}$$
- step9: Calculate:
  $$\begin{align}&\theta =\left\{ \begin{array}{l}\frac{\pi }{6}+k\pi \\frac{5\pi }{6}+k\pi \end{array}\right.,k \in \mathbb{Z}\&\cos\left(2\theta \right)=-\frac{1}{2}\end{align}$$
- step10: Calculate:
  $$\begin{align}&\theta =\left\{ \begin{array}{l}\frac{\pi }{6}+k\pi \\frac{5\pi }{6}+k\pi \end{array}\right.,k \in \mathbb{Z}\&\theta =\left\{ \begin{array}{l}\frac{\pi }{3}+k\pi \\frac{2\pi }{3}+k\pi \end{array}\right.,k \in \mathbb{Z}\end{align}$$
- step11: Find the union:
  $$\theta =\left\{ \begin{array}{l}\frac{\pi }{6}+\frac{k\pi }{2}\\frac{\pi }{3}+\frac{k\pi }{2}\end{array}\right.,k \in \mathbb{Z}$$
- step12: Check if the solution is in the defined range:
  $$\theta =\left\{ \begin{array}{l}\frac{\pi }{6}+\frac{k\pi }{2}\\frac{\pi }{3}+\frac{k\pi }{2}\end{array}\right.,k \in \mathbb{Z},\theta \neq \frac{\pi }{4}+\frac{k\pi }{2},k \in \mathbb{Z}$$
- step13: Find the intersection:
  $$\theta =\left\{ \begin{array}{l}\frac{\pi }{6}+\frac{k\pi }{2}\\frac{\pi }{3}+\frac{k\pi }{2}\end{array}\right.,k \in \mathbb{Z}$$
Solve the equation $$ \theta = \frac{1}{2} \arccos(0.866) - 5^{\text{o}} $$.
Solve the equation by following steps:
- step0: Solve for $$\theta$$:
  $$\theta =\frac{1}{2}\arccos\left(0.866\right)-5$$
- step1: Simplify:
  $$\theta =\frac{\arccos\left(0.866\right)-10}{2}$$
- step2: Simplify:
  $$\theta \approx -4.738175$$
Let's solve the equations step by step.

### 1) Solve the following equations:

#### a) $$ \sin \theta = 0.35 $$
To find $$ \theta $$:
$$
\theta = \arcsin(0.35) \approx 0.3576 \text{ radians} \approx 20.7^{\circ}
$$

#### b) $$ \tan 2\theta = 5 $$
To find $$ \theta $$:
$$
2\theta = \arctan(5) + k\pi, \quad k \in \mathbb{Z}
$$
Thus,
$$
\theta = \frac{\arctan(5)}{2} + \frac{k\pi}{2} \approx 0.6867 + \frac{k\pi}{2}, \quad k \in \mathbb{Z}
$$

#### c) $$ \cos(2\theta + 10^{\circ}) = 0.866 $$
To find $$ \theta $$:
$$
2\theta + 10^{\circ} = \arccos(0.866) + k \cdot 360^{\circ}, \quad k \in \mathbb{Z}
$$
Calculating $$ \arccos(0.866) $$:
$$
2\theta = 60^{\circ} - 10^{\circ} + k \cdot 360^{\circ} \implies 2\theta = 50^{\circ} + k \cdot 360^{\circ}
$$
Thus,
$$
\theta = 25^{\circ} + \frac{k \cdot 360^{\circ}}{2}, \quad k \in \mathbb{Z}
$$

#### d) $$ \cos(x + 20^{\circ}) = \sin 70^{\circ} $$
Using the identity $$ \sin 70^{\circ} = \cos 20^{\circ} $$:
$$
\cos(x + 20^{\circ}) = \cos 20^{\circ}
$$
This gives:
$$
x + 20^{\circ} = 20^{\circ} + k \cdot 360^{\circ} \quad \text{or} \quad x + 20^{\circ} = -20^{\circ} + k \cdot 360^{\circ}
$$
Thus,
$$
x = k \cdot 360^{\circ} \quad \text{or} \quad x = -40^{\circ} + k \cdot 360^{\circ}, \quad k \in \mathbb{Z}
$$

### 2) Solve the following equations:

#### a) $$ \sin^2 x - \sin x \cos x = 0 $$
Factoring gives:
$$
\sin x (\sin x - \cos x) = 0
$$
Thus,
$$
\sin x = 0 \quad \text{or} \quad \sin x = \cos x \implies x = k\pi \quad \text{or} \quad x = \frac{\pi}{4} + k\pi, \quad k \in \mathbb{Z}
$$

#### b) $$ 2 \sin x \tan x - \tan x = 0 $$
Factoring gives:
$$
\tan x (2 \sin x - 1) = 0
$$
Thus,
$$
\tan x = 0 \quad \text{or} \quad 2 \sin x - 1 = 0 \implies x = k\pi \quad \text{or} \quad x = \frac{\pi}{6} + k\pi, \quad k \in \mathbb{Z}
$$

#### c) $$ 8 \cos 2\theta = \frac{2}{\cos 2\theta} $$
Multiplying both sides by $$ \cos 2\theta $$:
$$
8 \cos^2 2\theta = 2 \implies 4 \cos^2 2\theta - 1 = 0
$$
Thus,
$$
\cos 2\theta = \pm \frac{1}{2} \implies 2\theta = 60^{\circ} + k \cdot 360^{\circ} \quad \text{or} \quad 2\theta = 120^{\circ} + k \cdot 360^{\circ}
$$
Thus,
$$
\theta = 30^{\circ} + k \cdot 180^{\circ} \quad \text{or} \quad \theta = 60^{\circ} + k \cdot 180^{\circ}, \quad k \in \mathbb{Z}
$$

#### d) $$ \sin x \cos x - 3 \cos x = 0 $$
Factoring gives:
$$
\cos x (\sin x - 3) = 0
$$
Thus,
$$
\cos x = 0 \quad \text{or} \quad \sin x = 3 \quad \text{(not possible)}
$$
Thus,
$$
x = \frac{\pi}{2} + k\pi, \quad k \in \mathbb{Z}
$$

#### e) $$ \sin(\theta - 20^{\circ}) = \cos \theta $$
Using the identity $$ \sin(\theta - 20^{\circ}) = \sin(90^{\circ} - \theta) $$:
$$
\theta - 20^{\circ} = 90^{\circ} - \theta + k \cdot 360^{\circ}
$$
Thus,
$$
2\theta = 110^{\circ} + k \cdot 360^{\circ} \implies \theta = 55^{\circ} + k \cdot 180^{\circ}, \quad k \in \mathbb{Z}
$$

### Summary of Results:
1. a) $$ \theta \approx 20.7^{\circ} $$
   b) $$ \theta \approx 0.6867 + \frac{k\pi}{2} $$
   c) $$ \theta = 25^{\circ} + k \cdot 180^{\circ} $$
   d) $$ x = k \cdot 360^{\circ} $$ or $$ x = -40^{\circ} + k \cdot 360^{\circ} $$

2. a) $$ x = k\pi $$ or $$ x = \frac{\pi}{4} + k\pi $$
   b) $$ x = k\pi $$ or $$ x = \frac{\pi}{6} + k\pi $$
   c) $$ \theta = 30^{\circ} + k \cdot 180^{\circ} $$ or $$ \theta = 60^{
1. a) $$ \theta \approx 20.7^\circ $$ b) $$ \theta \approx 0.6867 + \frac{k\pi}{2} $$ c) $$ \theta = 25^\circ + k \cdot 180^\circ $$ d) $$ x = k \cdot 360^\circ $$ or $$ x = -40^\circ + k \cdot 360^\circ $$ 2. a) $$ x = k\pi $$ or $$ x = \frac{\pi}{4} + k\pi $$ b) $$ x = k\pi $$ or $$ x = \frac{\pi}{6} + k\pi $$ c) $$ \theta = 30^\circ + k \cdot 180^\circ $$ or $$ \theta = 60^\circ + k \cdot 180^\circ $$ d) $$ \theta = 55^\circ + k \cdot 180^\circ $$

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