Question

Sin^2x+6sinxsinx/2+9=9cos^2x/2

Answer 1

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Answer 2

I can help you with your math problem. Here are the steps to solve it:

- First, use the double angle formula to rewrite the equation as $$\sin^2x+6\sin x\cos x+9=9(1-\sin^2x)$$ - Next, simplify and rearrange the equation to get $$10\sin^2x+6\sin x-9=0$$ - Then, factor the quadratic equation to get $$(5\sin x-3)(2\sin x+3)=0$$ - Finally, solve for x by setting each factor to zero and using the inverse sine function. The solutions are $$x=\arcsin\left(\frac{3}{5}\right)+2k\pi$$ or $$x=\arcsin\left(-\frac{3}{2}\right)+2k\pi$$ where k is any integer.

You can also check out these websites for more examples and explanations of trigonometric equations, , . I hope this helps. Have a nice day!