\cos(2x)+2\sin^2(x)>0
error msg
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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
More Steps
Evaluate
\cos\left(2x\right)+2\sin^{2}\left(x\right)
Transform the expression
More Steps
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)
The sum of two opposites equals 0
More Steps
Evaluate
\sin^{2}\left(x\right)-\sin^{2}\left(x\right)
Collect like terms
\left(1-1\right)\sin^{2}\left(x\right)
Add the coefficients
0\times \sin^{2}\left(x\right)
Calculate
0
1+0
Remove 0
1
1>0
Check the inequality
\textrm{true}
Solution
x \in \mathbb{R}
Alternative Form
\textrm{All real solution}
Show Solutions