Solve for - if 4 cos 2-2 sin-4-0- where 0 - 90

The correct options are A 0^∘ B 30^∘4cos^2θ + 2sinθ – 4 = 0 4[1 sin^2θ] + 2sinθ – 4 = 0 4 4sin^2θ + 2sinθ – 4 = 0 2sinθ [1 2sinθ] = 0 sinθ = 0 or sinθ = ...

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Byju's Answer

Standard X

Mathematics

Trigonometric Identities

Solve for θ i...

Question

Solve for $θ$ if $4 c o s^{2} θ$ + 2 $s i n θ$ – 4 = 0, where 0 $\leq$ $θ$ $\leq$ 90.

0^{\circ}

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30^{\circ}

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60^{\circ}

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45^{\circ}

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Solution

The correct options are
A $0^{\circ}$
B $30^{\circ}$
$4 c o s^{2} θ$ + 2 $s i n θ$ – 4 = 0

4[1 - $s i n^{2} θ$ ] + 2 $s i n θ$ – 4 = 0

4 - 4 $s i n^{2} θ$ + 2 $s i n θ$ – 4 = 0

2 $s i n θ$ [1 - 2 $s i n θ$ ] = 0

$s i n θ$ = 0 or $s i n θ$ = $\frac{1}{2}$
So, $θ = 0^{\circ} o r θ = 30^{\circ}$

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Similar questions

Solve for $θ$ if $4 c o s^{2} θ$ + 2 $s i n θ$ – 4 = 0, where $0^{\circ}$ $\leq$ $θ$ $\leq$ $90^{\circ}$ .

Q. Solve for

θ

4 {cos}^{2} θ

+ 2

sin θ

– 4 = 0, where

0^{\circ}

\leq

θ

\leq

90^{\circ}

Q. Solve

1 - 2 sin θ - 2 cos θ + cot θ = 0

(0 < θ < 2 π)

$3 tan θ$ + $cot θ$ = 5cosec $θ$ . Solve for $θ$ , $0 \leq θ \leq 90$ .

Q. Solve for

0^{\circ} < θ < 90^{\circ}

2 cos 3 θ = 1

2 sin 2 θ = \sqrt{3}

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Solve for θ if 4cos2θ + 2sinθ – 4 = 0, where 0 ≤ θ ≤ 90.

The correct options are A 0∘ B 30∘4cos2θ + 2sinθ – 4 = 0 4[1 - sin2θ] + 2sinθ – 4 = 0 4 - 4sin2θ + 2sinθ – 4 = 0 2sinθ [1 - 2sinθ] = 0 sinθ = 0 or sinθ = 12 So, θ=0∘ or θ=30∘

Solve for $θ$ if $4 c o s^{2} θ$ + 2 $s i n θ$ – 4 = 0, where 0 $\leq$ $θ$ $\leq$ 90.