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Solve each equation algebraically over the domain \(0 \leq x<2 \pi\). a) \(\tan ^{2} x-\tan x=0\) b) \(\sin 2 x-\sin x=0\) c) \(\sin ^{2} x-4 \sin x=5\) d) \(\cos 2 x=\sin x\)

Short Answer

Expert verified
a) \(x = 0, \frac{\pi}{4}, \pi, \frac{5\pi}{4}\); b) \(x = 0, \pi, \frac{\pi}{3}, \frac{5\pi}{3}\); c) \(x = \frac{3\pi}{2}\); d) \(x = \frac{\pi}{6}, \frac{5\pi}{6}, \frac{3\pi}{2}\)

Step by step solution

01

Solve Equation a) \( \tan ^{2} x - \tan x = 0 \)

Factor out \( \tan x \) from the left side of the equation: \( \tan x (\tan x - 1) = 0 \). This gives us two solutions: \( \tan x = 0 \) and \( \tan x - 1 = 0 \). Solving these, we get \( \tan x = 0 \) and \( \tan x = 1 \).
02

Find solutions for \( \tan x = 0 \)

\( \tan x = 0 \) when \( x = 0 \) or \( x = \pi \) within the given domain \( 0 \leq x < 2\pi \).
03

Find solutions for \( \tan x = 1 \)

\( \tan x = 1 \) when \( x = \frac{\pi}{4} \) or \( x = \frac{5\pi}{4} \) within the given domain \( 0 \leq x < 2\pi \). Thus, solutions for part a) are \( x = 0, \frac{\pi}{4}, \pi, \frac{5\pi}{4} \).
04

Solve Equation b) \( \sin 2 x - \sin x = 0 \)

Rewrite \( \sin 2x \) as \( 2 \sin x \cos x \): \( 2 \sin x \cos x - \sin x = 0 \). Factor out \( \sin x \): \( \sin x (2 \cos x - 1) = 0 \). This gives us two solutions: \( \sin x = 0 \) and \( 2 \cos x - 1 = 0 \).
05

Find solutions for \( \sin x = 0 \)

\( \sin x = 0 \) when \( x = 0, \pi, 2\pi \). Since \( x < 2\pi \), the valid solutions within the domain are \( x = 0 \) and \( x = \pi \).
06

Find solutions for \( 2 \cos x - 1 = 0 \)

\( 2 \cos x - 1 = 0 \) leads to \( \cos x = \frac{1}{2} \) when \( x = \frac{\pi}{3} \) or \( x = \frac{5 \pi}{3} \). Thus, solutions for part b) are \( x = 0, \pi, \frac{\pi}{3}, \frac{5\pi}{3} \).
07

Solve Equation c) \( \sin ^{2} x - 4 \sin x = 5 \)

Let \( y = \sin x \). The equation becomes \( y^2 - 4y - 5 = 0 \). Factorize it: \( (y - 5)(y + 1) = 0 \). This gives \( y = 5 \) and \( y = -1 \).
08

Solve for \( \sin x = 5 \) and \( \sin x = -1 \)

\( \sin x = 5 \) has no solutions as \( \sin x \) must be in \([-1, 1]\). \( \sin x = -1 \) when \( x = \frac{3\pi}{2} \). Thus, the solution for part c) is \( x = \frac{3 \pi}{2} \).
09

Solve Equation d) \( \cos 2 x = \sin x \)

Rewrite \( \cos 2x \) using its identity: \( \cos 2x = 1 - 2 \sin^2 x \). The equation becomes \( 1 - 2 \sin^2 x = \sin x \). Rearrange it to form a quadratic: \( 2 \sin^2 x + \sin x - 1 = 0 \).
10

Solve quadratic equation \( 2 \sin^2 x + \sin x - 1 = 0 \)

Use the quadratic formula \( y = \sin x \): \( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). Here, \( a = 2 \), \( b = 1 \), and \( c = -1 \). \( y = \frac{-1 \pm \sqrt{1 + 8}}{4} = \frac{-1 \pm 3}{4} \). So, \( y = \frac{1}{2} \) and \( y = -1 \).
11

Find solutions for \( \sin x = \frac{1}{2} \) and \( \sin x = -1 \)

\( \sin x = \frac{1}{2} \) when \( x = \frac{\pi}{6} \) or \( x = \frac{5 \pi}{6} \). From earlier, \( \sin x = -1 \) when \( x = \frac{3 \pi}{2} \). Thus, solutions for part d) are \( x = \frac{\pi}{6}, \frac{5\pi}{6}, \frac{3 \pi}{2} \).

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Key Concepts

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Trigonometric Identities
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