Question 775657: solve 2sinx+cosx=0 Answer by Tatiana_Stebko(1539) (Show Source):
To solve a trigonometric simplify the equation using trigonometric identities. Then, write the equation in a standard form, and isolate the variable using algebraic manipulation to solve for …
\sin (x)+\sin (\frac{x}{2})=0,\:0\le \:x\le \:2\pi \cos (x)-\sin (x)=0 \sin (4\theta)-\frac{\sqrt{3}}{2}=0,\:\forall 0\le\theta<2\pi ; 2\sin ^2(x)+3=7\sin (x),\:x\in[0,\:2\pi ] 3\tan ^3(A) …
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sin(x)(1+ 2cos(x)) = 0 sin (x) (1 + 2 cos (x)) = 0. If any individual factor on the left side of the equation is equal to 0 0, the entire expression will be equal to 0 0. sin(x) = 0 sin (x) = 0. …
Take the inversecosine of both sides of the equation to extract from inside the cosine. Simplify the right side. Tap for more steps. The exact value of is . The cosinefunction is positive in the first …
f(x) = sin x(2cos x + 1 ) = 0. Next, use the trig unit circle and trig table to solve the 2 basic trig equations: sin x = 0, and (2cos x + 1) = 0. sin x = 0 --> x = 0; x = Pi; x = 2Pi (within …
How do you solve sin 2x − cos x = 0? Use the important double angle identity sin2x = 2sinxcosx to start the solving process. 2sinxcosx − cosx = 0. cosx(2sinx − 1) = 0. cosx = 0 or …
How do you solve \displaystyle{\sin{{x}}}-{2}{\sin{{x}}}{\cos{{x}}}={0} over the interval 0 to 2pi?
After getting 2sinxcosx= cosx, you cannot divide the both sides by cosx because cosx can be zero. So, we have 2sinxcosx−cosx = 0 cosx(2sinx−1)= 0 cosx = 0 or 2sinx−1 = 0. Which …
Solving \cos(2x)-\sin(x)= 0 within the domain [0,2\pi]. Why am I missing some solutions?
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Transform the equation into 2 basic trig equations: 2sin x (cos x - 1) = 0. Next, solve the 2 basic equations: sin x = 0, and cos x = 1. Transformation process. There are 2 main approaches to …
Factor sin(x) sin (x) out of 2sin(x)cos(x)−sin(x) 2 sin (x) cos (x) - sin (x). Tap for more steps. sin(x)(2cos(x)−1) = 0 sin (x) (2 cos (x) - 1) = 0. If any individual factor on the left side of the …
x^{2}-x-6=0 -x+3\gt 2x+1 ; line\:(1,\:2),\:(3,\:1) f(x)=x^3 ; prove\:\tan^2(x)-\sin^2(x)=\tan^2(x)\sin^2(x) \frac{d}{dx}(\frac{3x+9}{2-x}) (\sin^2(\theta))' \sin(120)
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