Question

if 2cos^2theta sin theta and 0°<=theta<=90° find the value of theta​

Answer 1

Answer:

$Please see the explanation.</p><p></p><p>Explanation:</p><p></p><p>Substitute 1−sin2(θ) for cos2(θ):</p><p></p><p>2(1−sin2(θ))+3sin(θ)=0</p><p></p><p>Use the distributive property:</p><p></p><p>2−2sin2(θ))+3sin(θ)=0</p><p></p><p>Multiply both side by -1:</p><p></p><p>2sin2(θ))−3sin(θ)−2=0</p><p></p><p>This is a quadratic where the variable is sin(θ).</p><p></p><p>It looks like it will factor:</p><p></p><p>#(sin(theta) - 2)(2sin(theta) + 1) = 0</p><p></p><p>sin(θ)=2andsin(θ)=−12</p><p></p><p>We must discard the first root, because it is outside the range of the sine function.</p><p></p><p>Turning our attention to the second root:</p><p></p><p>sin(θ)=−12</p><p></p><p>Rotating counterclockwise from 0, the first encounter of this is at:</p><p></p><p>θ=7π6</p><p></p><p>The next encounter with this is at:</p><p></p><p>θ=11π6</p><p></p><p>Add integer rotations of 2π to both:</p><p></p><p>θ=7π6+2nπ and θ=11π6+2nπ Where n can be any integer (positive, negative, or zero)</p><p></p><p>$

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Answer 2

Answer:

it's 30°

Step-by-step explanation:

hope it's helpful