Answer :
Sure, let's work through each problem step-by-step.
### Problem 1: Solving [tex]\( 1-2 \cos \left(\theta-\frac{\pi}{5}\right)=0 \)[/tex] for [tex]\( -\pi<\theta \leqslant \pi \)[/tex]
First, rewrite the equation:
[tex]\[ 1 - 2 \cos \left(\theta - \frac{\pi}{5}\right) = 0 \][/tex]
Subtract 1 from both sides:
[tex]\[ -2 \cos \left(\theta - \frac{\pi}{5}\right) = -1 \][/tex]
Divide both sides by -2:
[tex]\[ \cos \left(\theta - \frac{\pi}{5}\right) = \frac{1}{2} \][/tex]
We need to find the values of [tex]\(\theta\)[/tex] that satisfy this equation. Recall that [tex]\(\cos \phi = \frac{1}{2}\)[/tex] has solutions:
[tex]\[ \phi = \frac{\pi}{3} \quad \text{or} \quad \phi = -\frac{\pi}{3} + 2k\pi \quad \text{for integer } k \][/tex]
Let's denote [tex]\(\phi = \theta - \frac{\pi}{5}\)[/tex]:
[tex]\[ \theta - \frac{\pi}{5} = \frac{\pi}{3} \quad \text{or} \quad \theta - \frac{\pi}{5} = -\frac{\pi}{3} + 2k\pi \][/tex]
Solve for [tex]\(\theta\)[/tex]:
[tex]\[ \theta = \frac{\pi}{3} + \frac{\pi}{5} = \frac{5\pi + 3\pi}{15} = \frac{8\pi}{15} \][/tex]
[tex]\[ \theta = -\frac{\pi}{3} + \frac{\pi}{5} + 2k\pi = -\frac{5\pi - 3\pi}{15} + 2k\pi = -\frac{2\pi}{15} + 2k\pi \][/tex]
We seek solutions where [tex]\(-\pi < \theta \leq \pi\)[/tex].
For [tex]\(k=0\)[/tex]:
[tex]\[ \theta = -\frac{2\pi}{15} \][/tex]
This is within [tex]\(-\pi\)[/tex] to [tex]\(\pi\)[/tex].
For [tex]\(k=1\)[/tex]:
[tex]\[ \theta = -\frac{2\pi}{15} + 2\pi = \frac{28\pi}{15} \quad \text{(not within the interval \(-\pi\) to \(\pi\))} \][/tex]
Thus, the solutions to [tex]\( \theta \)[/tex] in the range [tex]\(-\pi < \theta \leq \pi\)[/tex] are:
[tex]\[ \theta = \frac{8\pi}{15} \quad \text{and} \quad \theta = -\frac{2\pi}{15}. \][/tex]
### Problem 2: Solving [tex]\( 4 \cos ^2 x+7 \sin x-2=0 \)[/tex] for [tex]\( 0 \leqslant x < 360^\circ \)[/tex]
First, use the Pythagorean identity [tex]\( \cos^2 x = 1 - \sin^2 x \)[/tex]:
[tex]\[ 4 (1 - \sin^2 x) + 7 \sin x - 2 = 0 \][/tex]
Simplify:
[tex]\[ 4 - 4 \sin^2 x + 7 \sin x - 2 = 0 \][/tex]
Rearrange terms:
[tex]\[ -4 \sin^2 x + 7 \sin x + 2 = 0 \][/tex]
Multiply through by -1 to make the quadratic equation standard:
[tex]\[ 4 \sin^2 x - 7 \sin x - 2 = 0 \][/tex]
Let [tex]\( y = \sin x \)[/tex]. Then we have a quadratic equation:
[tex]\[ 4y^2 - 7y - 2 = 0 \][/tex]
We solve this quadratic equation using the quadratic formula, [tex]\( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex]:
Here, [tex]\( a = 4 \)[/tex], [tex]\( b = -7 \)[/tex], [tex]\( c = -2 \)[/tex].
[tex]\[ y = \frac{7 \pm \sqrt{(-7)^2 - 4 \cdot 4 \cdot (-2)}}{2 \cdot 4} \][/tex]
[tex]\[ y = \frac{7 \pm \sqrt{49 + 32}}{8} \][/tex]
[tex]\[ y = \frac{7 \pm \sqrt{81}}{8} \][/tex]
[tex]\[ y = \frac{7 \pm 9}{8} \][/tex]
So, we have two possible values for [tex]\( y \)[/tex]:
[tex]\[ y = \frac{16}{8} = 2 \quad \text{(discard, since } -1 \leq y \leq 1) \][/tex]
[tex]\[ y = \frac{-2}{8} = -\frac{1}{4} \][/tex]
Thus, [tex]\(\sin x = -\frac{1}{4}\)[/tex].
Using the arcsine function:
[tex]\[ x = \arcsin(-\frac{1}{4}) \][/tex]
For [tex]\( 0^\circ \leqslant x < 360^\circ \)[/tex], [tex]\(\sin x = -\frac{1}{4}\)[/tex] occurs in the third and fourth quadrants:
[tex]\[ x = 180^\circ + \arcsin(-\frac{1}{4}) \quad \text{and} \quad x = 360^\circ - \arcsin(-\frac{1}{4}) \][/tex]
Calculating these, we use a calculator:
[tex]\[ \arcsin(-\frac{1}{4}) \approx -14.5^\circ \][/tex]
[tex]\[ x \approx 180^\circ - 14.5^\circ \quad \Rightarrow \quad x \approx 165.5^\circ \][/tex]
[tex]\[ x \approx 360^\circ + 14.5^\circ \quad \Rightarrow \quad x \approx 345.5^\circ \][/tex]
Thus, the solutions in the range [tex]\(0^\circ \leq x < 360^\circ\)[/tex] are:
[tex]\[ x \approx 165.5^\circ \quad \text{and} \quad x \approx 345.5^\circ \][/tex]
### Problem 1: Solving [tex]\( 1-2 \cos \left(\theta-\frac{\pi}{5}\right)=0 \)[/tex] for [tex]\( -\pi<\theta \leqslant \pi \)[/tex]
First, rewrite the equation:
[tex]\[ 1 - 2 \cos \left(\theta - \frac{\pi}{5}\right) = 0 \][/tex]
Subtract 1 from both sides:
[tex]\[ -2 \cos \left(\theta - \frac{\pi}{5}\right) = -1 \][/tex]
Divide both sides by -2:
[tex]\[ \cos \left(\theta - \frac{\pi}{5}\right) = \frac{1}{2} \][/tex]
We need to find the values of [tex]\(\theta\)[/tex] that satisfy this equation. Recall that [tex]\(\cos \phi = \frac{1}{2}\)[/tex] has solutions:
[tex]\[ \phi = \frac{\pi}{3} \quad \text{or} \quad \phi = -\frac{\pi}{3} + 2k\pi \quad \text{for integer } k \][/tex]
Let's denote [tex]\(\phi = \theta - \frac{\pi}{5}\)[/tex]:
[tex]\[ \theta - \frac{\pi}{5} = \frac{\pi}{3} \quad \text{or} \quad \theta - \frac{\pi}{5} = -\frac{\pi}{3} + 2k\pi \][/tex]
Solve for [tex]\(\theta\)[/tex]:
[tex]\[ \theta = \frac{\pi}{3} + \frac{\pi}{5} = \frac{5\pi + 3\pi}{15} = \frac{8\pi}{15} \][/tex]
[tex]\[ \theta = -\frac{\pi}{3} + \frac{\pi}{5} + 2k\pi = -\frac{5\pi - 3\pi}{15} + 2k\pi = -\frac{2\pi}{15} + 2k\pi \][/tex]
We seek solutions where [tex]\(-\pi < \theta \leq \pi\)[/tex].
For [tex]\(k=0\)[/tex]:
[tex]\[ \theta = -\frac{2\pi}{15} \][/tex]
This is within [tex]\(-\pi\)[/tex] to [tex]\(\pi\)[/tex].
For [tex]\(k=1\)[/tex]:
[tex]\[ \theta = -\frac{2\pi}{15} + 2\pi = \frac{28\pi}{15} \quad \text{(not within the interval \(-\pi\) to \(\pi\))} \][/tex]
Thus, the solutions to [tex]\( \theta \)[/tex] in the range [tex]\(-\pi < \theta \leq \pi\)[/tex] are:
[tex]\[ \theta = \frac{8\pi}{15} \quad \text{and} \quad \theta = -\frac{2\pi}{15}. \][/tex]
### Problem 2: Solving [tex]\( 4 \cos ^2 x+7 \sin x-2=0 \)[/tex] for [tex]\( 0 \leqslant x < 360^\circ \)[/tex]
First, use the Pythagorean identity [tex]\( \cos^2 x = 1 - \sin^2 x \)[/tex]:
[tex]\[ 4 (1 - \sin^2 x) + 7 \sin x - 2 = 0 \][/tex]
Simplify:
[tex]\[ 4 - 4 \sin^2 x + 7 \sin x - 2 = 0 \][/tex]
Rearrange terms:
[tex]\[ -4 \sin^2 x + 7 \sin x + 2 = 0 \][/tex]
Multiply through by -1 to make the quadratic equation standard:
[tex]\[ 4 \sin^2 x - 7 \sin x - 2 = 0 \][/tex]
Let [tex]\( y = \sin x \)[/tex]. Then we have a quadratic equation:
[tex]\[ 4y^2 - 7y - 2 = 0 \][/tex]
We solve this quadratic equation using the quadratic formula, [tex]\( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex]:
Here, [tex]\( a = 4 \)[/tex], [tex]\( b = -7 \)[/tex], [tex]\( c = -2 \)[/tex].
[tex]\[ y = \frac{7 \pm \sqrt{(-7)^2 - 4 \cdot 4 \cdot (-2)}}{2 \cdot 4} \][/tex]
[tex]\[ y = \frac{7 \pm \sqrt{49 + 32}}{8} \][/tex]
[tex]\[ y = \frac{7 \pm \sqrt{81}}{8} \][/tex]
[tex]\[ y = \frac{7 \pm 9}{8} \][/tex]
So, we have two possible values for [tex]\( y \)[/tex]:
[tex]\[ y = \frac{16}{8} = 2 \quad \text{(discard, since } -1 \leq y \leq 1) \][/tex]
[tex]\[ y = \frac{-2}{8} = -\frac{1}{4} \][/tex]
Thus, [tex]\(\sin x = -\frac{1}{4}\)[/tex].
Using the arcsine function:
[tex]\[ x = \arcsin(-\frac{1}{4}) \][/tex]
For [tex]\( 0^\circ \leqslant x < 360^\circ \)[/tex], [tex]\(\sin x = -\frac{1}{4}\)[/tex] occurs in the third and fourth quadrants:
[tex]\[ x = 180^\circ + \arcsin(-\frac{1}{4}) \quad \text{and} \quad x = 360^\circ - \arcsin(-\frac{1}{4}) \][/tex]
Calculating these, we use a calculator:
[tex]\[ \arcsin(-\frac{1}{4}) \approx -14.5^\circ \][/tex]
[tex]\[ x \approx 180^\circ - 14.5^\circ \quad \Rightarrow \quad x \approx 165.5^\circ \][/tex]
[tex]\[ x \approx 360^\circ + 14.5^\circ \quad \Rightarrow \quad x \approx 345.5^\circ \][/tex]
Thus, the solutions in the range [tex]\(0^\circ \leq x < 360^\circ\)[/tex] are:
[tex]\[ x \approx 165.5^\circ \quad \text{and} \quad x \approx 345.5^\circ \][/tex]
