Solved: "Solve the equations and inequalities. a)32x−4×3x+3=0 b)sin⁡θ=sin⁡2θ,0≤θ≤2π c)sin2θ−cos2θ=1+cos⁡θ,0≤θ≤2π"

Turnseeuw

Answered question

2022-01-24

Solve the equations and inequalities.
a)

3^{2 x} - 4 \times 3^{x} + 3 = 0

\sin θ = \sin 2 θ, 0 \leq θ \leq 2 π

\sin^{2} θ - \cos^{2} θ = 1 + \cos θ, 0 \leq θ \leq 2 π

trovabile4p

Beginner2022-01-25Added 13 answers

3^{2 x} - 4 \times 3^{x} + 3 = 0

\Rightarrow {(3^{x})}^{2} - 4 (3^{x}) + 3 = 0

\Rightarrow 3^{x} = \frac{4 \pm \sqrt{16 - 12}}{2}

= \frac{4 \pm \sqrt{4}}{2}

= \frac{4 \pm 2}{2}

\Rightarrow 3^{x} = 4 + 2

3^{x} = 4 - 2

\Rightarrow 3^{x} = 3

3^{x} = 1 = 3^{0}

\Rightarrow x = 1

x = 0

\sin θ = \sin 2 θ, 0 \leq θ \leq 2 π

∵ \sin (2 θ) = 2 \sin θ \cos θ

\Rightarrow \sin θ = 2 \sin θ \cos θ

\Rightarrow \sin θ (1 - 2 \cos θ) = 0

\Rightarrow \sin θ = 0

1 - 2 \cos θ = 0

\Rightarrow θ = 0, π, 2 π

\cos θ = \frac{1}{2}

\Rightarrow θ = \frac{π}{3}, \frac{5 π}{3}

when

0 \leq θ \leq 2 π

Karli Kaiser

Beginner2022-01-26Added 9 answers

(c)

\sin^{2} θ - \cos^{2} θ = 1 + \cos θ, 0 \leq θ \leq 2 π

We know that

\cos 2 θ = \cos^{2} θ - \sin^{2} θ

\cos 2 θ = 2 \cos^{2} θ - 1

\Rightarrow - \cos (2 θ) = 1 + \cos θ

\Rightarrow - (2 \cos^{2} θ - 1) = + \cos θ

\Rightarrow 2 \cos^{2} θ - 1 = - 1 - \cos θ

\Rightarrow 2 \cos^{2} θ + \cos θ = 0

\Rightarrow \cos θ (2 \cos θ + 1) = 0

\Rightarrow \cos θ = 0

\cos θ = \frac{- 1}{2}

\Rightarrow θ = \frac{π}{2}, \frac{3 π}{2}

θ = \frac{2 π}{3}, \frac{4 π}{3}

When

0 \leq θ \leq 2 π

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