3 sin ⁡ x + 1 = sin ⁡ x 3\sin\ x\ +\ 1\ =\ \sin\ x 3 sin x + 1 = sin x

x = 7 π 6 + 2 π n , x = 11 π 6 + 2 π n x=\frac{7\pi}{6}+2\pi n,\ \ \ x\ =\ \frac{11\pi}{6}+2\pi n x = 6 7 π + 2 π n , x = 6 1 1 π + 2 π n

x = π 6 + 2 π n , x = 5 π 6 + 2 π n x=\frac{\pi}{6}+2\pi n,\ \ \ x\ =\ \frac{5\pi}{6}+2\pi n x = 6 π + 2 π n , x = 6 5 π + 2 π n

x = 4 π 3 + 2 π n , x = 5 π 3 + 2 π n x=\frac{4\pi}{3}+2\pi n,\ \ \ \ x=\frac{5\pi}{3}+2\pi n x = 3 4 π + 2 π n , x = 3 5 π + 2 π n

x = π 4 + π n x=\frac{\pi}{4}+\pi n x = 4 π + π n

Solve equation for 0 ≤ θ < 2 π 0\le\theta<2\pi 0 ≤ θ < 2 π . 1 = 5 − 8 cos ⁡ θ 1=5-8\cos\theta 1 = 5 − 8 cos θ

θ = π 3 , 5 π 3 \theta=\frac{\pi}{3},\frac{5\pi}{3} θ = 3 π , 3 5 π

θ = π 6 , π 3 , 11 π 6 \theta=\frac{\pi}{6},\frac{\pi}{3},\frac{11\pi}{6} θ = 6 π , 3 π , 6 1 1 π

θ = π 6 , 11 π 6 \theta=\frac{\pi}{6},\frac{11\pi}{6} θ = 6 π , 6 1 1 π

θ = π 6 \theta=\frac{\pi}{6} θ = 6 π

Solve equation for 0 ≤ θ < 2 π 0\le\theta<2\pi 0 ≤ θ < 2 π . − 2 + cot ⁡ θ = − 3 -2+\cot\theta=-3 − 2 + cot θ = − 3

θ = 3 π 4 , 7 π 4 \theta=\frac{3\pi}{4},\frac{7\pi}{4} θ = 4 3 π , 4 7 π

θ = 3 π 4 , 4 π 3 , 7 π 4 \theta=\frac{3\pi}{4},\frac{4\pi}{3},\frac{7\pi}{4} θ = 4 3 π , 3 4 π , 4 7 π

θ = π 3 , 4 π 3 \theta=\frac{\pi}{3},\frac{4\pi}{3} θ = 3 π , 3 4 π

θ = 3 π 4 , 4 π 3 \theta=\frac{3\pi}{4},\frac{4\pi}{3} θ = 4 3 π , 3 4 π

Solve equation for 0 ≤ θ < 2 π 0\le\theta<2\pi 0 ≤ θ < 2 π . − sin ⁡ θ csc ⁡ θ + 3 csc ⁡ θ = 2 sin ⁡ θ + 3 csc ⁡ θ -\sin\theta\csc\theta+3\csc\theta=\sqrt{2}\sin\theta+3\csc\theta − sin θ csc θ + 3 csc θ = 2 <path d="M95,702c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5, -10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8, -50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0, 35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5, -221c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467 s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422 s-65,47,-65,47z M834 80H400000v40H845z"> sin θ + 3 csc θ

θ = 5 π 4 , 7 π 4 \theta=\frac{5\pi}{4},\frac{7\pi}{4} θ = 4 5 π , 4 7 π

θ = 5 π 4 , 11 π 6 \theta=\frac{5\pi}{4},\frac{11\pi}{6} θ = 4 5 π , 6 1 1 π

θ = 3 π 4 , 7 π 6 , 7 π 4 \theta=\frac{3\pi}{4},\frac{7\pi}{6},\frac{7\pi}{4} θ = 4 3 π , 6 7 π , 4 7 π

θ = 0 , π , 5 π 4 , 7 π 4 \theta=0,\pi,\frac{5\pi}{4},\frac{7\pi}{4} θ = 0 , π , 4 5 π , 4 7 π

Find the solution on the interval [0, 2π)

0; π; π/4; 3π/4; 5π/4; 7π/4

Find the solution over the interval [0, 2π)

0, π, π/6, 5π/6, 7π/6, 11π/6

Solving Trig Equations

Authored by Jennifer Gray

Mathematics

11th Grade

CCSS covered

Used 16+ times

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25 questions

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MULTIPLE CHOICE QUESTION

30 sec • 1 pt

Choose a good way to start solving this equation
tanx sin²x = 2 tanx

divide tanx from both sides

factor out tanx

subtract 2 tanx from the left and then factor tanx out

cancel sin²x out

Tags

CCSS.HSF.TF.B.7

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

Solve on the domain [0, 2π)
0 = sinx(sinx − 1)

0, π

0, π, ½π

½π

0, 90, 270

Tags

CCSS.HSF.TF.B.7

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

Which is a correct way to solve the equation
cosx(2cosx − 3) = 0?

divide cos x from both sides

set each factor equal to 0 and solve

distribute cosx into the parentheses

guess and hope for the best

Tags

CCSS.HSF.TF.B.7

MULTIPLE CHOICE QUESTION

5 mins • 1 pt

Solve on the domain [0, 2π)
tan(x)+1=2

0 and π

3π/4 and 7π/4

π/4 and 5π/4

3π/4 and 5π/4

Tags

CCSS.HSF.TF.B.7

MULTIPLE CHOICE QUESTION

3 mins • 1 pt

Solve on the domain [0, 2π)
2cos²θ -1 = 0

θ = π /4, 7π /4

θ = π /4, 3π /4

θ = 3π /4, 5π /4

θ = π /4, 3π /4, 5π /4, 7π /4

Tags

CCSS.HSF.TF.B.7

MULTIPLE CHOICE QUESTION

30 sec • 2 pts

Solve on the domain [0, 2π)
1/2sec x - 1 = 0

π/6, 5π/6

π/3, 5π/3

2π/3, 4π/3

7π/6, 11π/6

Tags

CCSS.HSF.TF.B.7

MULTIPLE CHOICE QUESTION

30 sec • 2 pts

Solve on the domain [0, 2π)
4sin²x = 3

π/6, 11π/6

π/3, 2π/3

π/6, 5π/6, 7π/6, 11π/6

π/3, 2π/3, 4π/3, 5π/3

Tags

CCSS.HSF.TF.B.7

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Solving Trig Equations

Choose a good way to start solving this equationtanx sin2x = 2 tanx

Solve on the domain [0, 2π)0 = sinx(sinx − 1)

Which is a correct way to solve the equationcosx(2cosx − 3) = 0?

Solve on the domain [0, 2π)tan(x)+1=2

Solve on the domain [0, 2π)2cos2θ -1 = 0

Solve on the domain [0, 2π)1/2sec x - 1 = 0

Solve on the domain [0, 2π)4sin2x = 3

Solve on the domain [0, 2π)2sin x cos x = √2 cos x

Access all questions and much more by creating a free account

Similar Resources on Wayground

Popular Resources on Wayground

Discover more resources for Mathematics

Choose a good way to start solving this equation
tanx sin²x = 2 tanx

Solve on the domain [0, 2π)
0 = sinx(sinx − 1)

Which is a correct way to solve the equation
cosx(2cosx − 3) = 0?

Solve on the domain [0, 2π)
tan(x)+1=2

Solve on the domain [0, 2π)
2cos²θ -1 = 0

Solve on the domain [0, 2π)
1/2sec x - 1 = 0

Solve on the domain [0, 2π)
4sin²x = 3

Solve on the domain [0, 2π)
2sin x cos x = √2 cos x