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\cos(2x)+2\sin^2(x)>0

Algebra Calculus Trigonometry Matrix

Question

\cos\left(2x\right)+2\sin^{2}\left(x\right)>0

Solve the inequality

x \in \mathbb{R}

Alternative Form

\textrm{All real solution}

Evaluate

\cos\left(2x\right)+2\sin^{2}\left(x\right)>0

Rewrite the expression

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Evaluate

\cos\left(2x\right)+2\sin^{2}\left(x\right)

Transform the expression

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Evaluate

2\sin^{2}\left(x\right)

\text{Use }\sin^2 t = 1 - \cos^2 t\text{ to transform the expression}

2\left(1-\cos^{2}\left(x\right)\right)

Use the the distributive property to expand the expression

2\times 1+2\left(-\cos^{2}\left(x\right)\right)

Any expression multiplied by 1 remains the same

2+2\left(-\cos^{2}\left(x\right)\right)

Multiply the terms

2-2\cos^{2}\left(x\right)

\cos\left(2x\right)+2-2\cos^{2}\left(x\right)

\text{Use }\cos(2t) = \cos^2t-\sin^2t\text{ to transform the expression}

\cos^{2}\left(x\right)-\sin^{2}\left(x\right)+2-2\cos^{2}\left(x\right)

Subtract the terms

-\cos^{2}\left(x\right)-\sin^{2}\left(x\right)+2

\text{Use }\cos^2 t = 1 - \sin^2 t\text{ to transform the expression}

-\left(1-\sin^{2}\left(x\right)\right)-\sin^{2}\left(x\right)+2

Calculate

-1+\sin^{2}\left(x\right)-\sin^{2}\left(x\right)+2

Add the numbers

1+\sin^{2}\left(x\right)-\sin^{2}\left(x\right)

Since two opposites add up to 0,remove them form the expression

1>0

Check the inequality

\textrm{true}

Solution

x \in \mathbb{R}

Alternative Form

\textrm{All real solution}

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