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Math  /  Trigonometry

Question7. Determine exact solutions for each equation in the interval x∈[0,2π]x \in[0,2 \pi]. a) sin⁡2x−14=0\sin ^{2} x-\frac{1}{4}=0 b) cos⁡2x−34=0\cos ^{2} x-\frac{3}{4}=0 c) tan⁡2x−3=0\tan ^{2} x-3=0 d) 3csc⁡2x−4=03 \csc ^{2} x-4=0 5.4 Solve Trigonometric Equations ⋅\cdot MHR 287

Studdy Solution

STEP 1

1. We are solving trigonometric equations within the interval x∈[0,2π] x \in [0, 2\pi] .
2. We will use trigonometric identities and algebraic manipulation to find the solutions.
3. The solutions should be exact and expressed in terms of radians.

STEP 2

1. Solve equation (a) for x x .
2. Solve equation (b) for x x .
3. Solve equation (c) for x x .
4. Solve equation (d) for x x .

STEP 3

For equation (a) sin⁡2x−14=0\sin^2 x - \frac{1}{4} = 0:
1. Add 14\frac{1}{4} to both sides to isolate sin⁡2x\sin^2 x:
$ \sin^2 x = \frac{1}{4} \]
2. Take the square root of both sides:
$ \sin x = \pm \frac{1}{2} \]
3. Find x x in the interval [0,2π][0, 2\pi] where sin⁡x=12\sin x = \frac{1}{2} and sin⁡x=−12\sin x = -\frac{1}{2}:
- sin⁡x=12\sin x = \frac{1}{2} gives x=π6,5π6x = \frac{\pi}{6}, \frac{5\pi}{6}. - sin⁡x=−12\sin x = -\frac{1}{2} gives x=7π6,11π6x = \frac{7\pi}{6}, \frac{11\pi}{6}.

STEP 4

For equation (b) cos⁡2x−34=0\cos^2 x - \frac{3}{4} = 0:
1. Add 34\frac{3}{4} to both sides to isolate cos⁡2x\cos^2 x:
$ \cos^2 x = \frac{3}{4} \]
2. Take the square root of both sides:
$ \cos x = \pm \frac{\sqrt{3}}{2} \]
3. Find x x in the interval [0,2π][0, 2\pi] where cos⁡x=32\cos x = \frac{\sqrt{3}}{2} and cos⁡x=−32\cos x = -\frac{\sqrt{3}}{2}:
- cos⁡x=32\cos x = \frac{\sqrt{3}}{2} gives x=0,11π6x = 0, \frac{11\pi}{6}. - cos⁡x=−32\cos x = -\frac{\sqrt{3}}{2} gives x=5π6,7π6x = \frac{5\pi}{6}, \frac{7\pi}{6}.

STEP 5

For equation (c) tan⁡2x−3=0\tan^2 x - 3 = 0:
1. Add 3 to both sides to isolate tan⁡2x\tan^2 x:
$ \tan^2 x = 3 \]
2. Take the square root of both sides:
$ \tan x = \pm \sqrt{3} \]
3. Find x x in the interval [0,2π][0, 2\pi] where tan⁡x=3\tan x = \sqrt{3} and tan⁡x=−3\tan x = -\sqrt{3}:
- tan⁡x=3\tan x = \sqrt{3} gives x=π3,4π3x = \frac{\pi}{3}, \frac{4\pi}{3}. - tan⁡x=−3\tan x = -\sqrt{3} gives x=2π3,5π3x = \frac{2\pi}{3}, \frac{5\pi}{3}.

STEP 6

For equation (d) 3csc⁡2x−4=03 \csc^2 x - 4 = 0:
1. Add 4 to both sides and divide by 3 to isolate csc⁡2x\csc^2 x:
$ \csc^2 x = \frac{4}{3} \]
2. Take the reciprocal and square root of both sides to find sin⁡x\sin x:
\sin^2 x = \frac{3}{4} \] \sin x = \pm \frac{\sqrt{3}}{2} \]
3. Find x x in the interval [0,2π][0, 2\pi] where sin⁡x=32\sin x = \frac{\sqrt{3}}{2} and sin⁡x=−32\sin x = -\frac{\sqrt{3}}{2}:
- sin⁡x=32\sin x = \frac{\sqrt{3}}{2} gives x=π3,2π3x = \frac{\pi}{3}, \frac{2\pi}{3}. - sin⁡x=−32\sin x = -\frac{\sqrt{3}}{2} gives x=4π3,5π3x = \frac{4\pi}{3}, \frac{5\pi}{3}.
The solutions for each equation are:
a) x=π6,5π6,7π6,11π6 x = \frac{\pi}{6}, \frac{5\pi}{6}, \frac{7\pi}{6}, \frac{11\pi}{6}
b) x=0,11π6,5π6,7π6 x = 0, \frac{11\pi}{6}, \frac{5\pi}{6}, \frac{7\pi}{6}
c) x=π3,4π3,2π3,5π3 x = \frac{\pi}{3}, \frac{4\pi}{3}, \frac{2\pi}{3}, \frac{5\pi}{3}
d) x=π3,2π3,4π3,5π3 x = \frac{\pi}{3}, \frac{2\pi}{3}, \frac{4\pi}{3}, \frac{5\pi}{3}

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